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{"title":"Capítulo 3: Igualdades exponenciais e logarítmicas","markdown":{"headingText":"Capítulo 3: Igualdades exponenciais e logarítmicas","headingAttr":{"id":"","classes":["unnumbered"],"keyvalue":[]},"containsRefs":false,"markdown":"\n```{=html}\n\n<div id=\"conteudo-capitulo\">\n\n```\n\n:::{.raw_html}\n \n <br />\n <br />\n <br />\n <p class=\" unidade\" id=\"3P1\" title=\"3P1\">\n Neste capítulo, explicaremos porque são válidas as identidades exponenciais mencionadas na introdução deste texto. As\n igualdades que usualmente definem as funções trigonométricas hiperbólicas, como soma de exponenciais,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P2\" title=\"3P2\"><a id=\"idexpo\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm{senh}}x = \\frac{e^{x}-e^{-x}}{2} \\qquad \\text{e} \\qquad \\cosh x = \\frac{e^{x} + e^{-x}}{2}.\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img10.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}x = \\frac{e^{x}-e^{-x}}{2}$\" loading=\"lazy\"> e<img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img11.svg\" alt=\"$\\displaystyle \\qquad \\cosh x = \\frac{e^{x} + e^{-x}}{2}. $\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">1</span>) </td></tr>\n </tbody></table></div>\n\n\n <p class=\" unidade\" id=\"3P3\" title=\"3P3\">\n Como sabemos, o logaritmo é a função inversa da função exponencial, com a devida restrição no domínio e na imagem. É\n natural então pensarmos que as funções trigonométricas hiperbólicas inversas possam ser escritas como logaritmos. Isto\n de fato ocorre e também mostraremos como são obtidas as fórmulas\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P4\" title=\"3P4\"><a id=\"idlog\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm{senh}}^{-1} x = \\ln(x + \\sqrt{x^{2} + 1}) \\qquad \\text{e} \\qquad \\cosh^{-1} x = \\ln(x + \\sqrt{x^{2} - 1}).\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1114.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}^{-1} x = \\ln(x + \\sqrt{x^{2} + 1})$\" loading=\"lazy\"> e<img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1115.svg\" alt=\"$\\displaystyle \\qquad \\cosh^{-1} x = \\ln(x + \\sqrt{x^{2} - 1}).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">2</span>) </td></tr>\n </tbody></table></div>\n\n\n <p class=\" unidade\" id=\"3P5\" title=\"3P5\">\n Identidades similares às identidades em (<a href=\"#idexpo\">3.1</a>) e em (<a href=\"#idlog\">3.2</a>) também são válidas para a trigonometria\n circular. Entretanto neste caso será necessário o envolvimento de variáveis complexas.\n </p>\n <a id=\"secspot\"></a>\n\n:::\n\n## 3.1 Método das séries de potência {#SECTION00710000000000000000}\n\n::: {.raw_html}\n\n\n <p class=\" unidade\" id=\"3P6\" title=\"3P6\">\n O método que utilizaremos para provar as identidades em (<a href=\"#idexpo\">3.1</a>) nesta seção é o método das séries de potência e\n então faremos primeiramente uma breve introdução às séries de potência de variáveis reais. Para um estudo mais\n aprofundado sobre séries de potências recomendamos [<a href=\"/trigonometria-hiperbolica/referencias#Swokowski\">8</a>, Swokowski].\n \n </p>\n <div><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">1</span></b> \n Se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> é uma variável real independente, então uma série de potências em <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> é uma soma infinita da forma,\n <!-- MATH\n \\begin{displaymath}\n \\sum_{n=0}^{\\infty} a_{n} x^{n} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + a_{5}x^{5} + \\cdots\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P7\" title=\"3P7\">\n <img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1116.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} a_{n} x^{n} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + a_{5}x^{5} + \\cdots $\" loading=\"lazy\">\n </div>\n sendo que cada termo <span class=\"MATH\"><img style=\"height: 1.48ex; vertical-align: -0.41ex; \" src=\"img/img1117.svg\" alt=\"$a_{n}$\" loading=\"lazy\"></span> é um número real, chamado de coeficiente da <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span>-ésima potência de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>.\n </div>\n \n <p class=\" unidade\" id=\"3P8\" title=\"3P8\">\n Observe que o primeiro termo da série é <!-- MATH\n $a_{0}x^{0}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.46ex; \" src=\"img/img1119.svg\" alt=\"$a_{0}x^{0}$\" loading=\"lazy\"></span>. Para simplificar a notação, estamos supondo que <span class=\"MATH\"><img style=\"height: 2.03ex; vertical-align: -0.11ex; \" src=\"img/img1120.svg\" alt=\"$x^{0} = 1$\" loading=\"lazy\"></span> mesmo\n para <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1121.svg\" alt=\"$x=0$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P9\" title=\"3P9\">\n Se a soma infinita existir e for um número real <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img7.svg\" alt=\"$S$\" loading=\"lazy\"></span>, então dizemos que a série converge, ou ainda, que converge para\n <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img7.svg\" alt=\"$S$\" loading=\"lazy\"></span>. Se a soma não existir então a série é dita divergente. Naturalmente a convergência de uma série de potências está\n condicionada aos termos <span class=\"MATH\"><img style=\"height: 1.48ex; vertical-align: -0.41ex; \" src=\"img/img1117.svg\" alt=\"$a_{n}$\" loading=\"lazy\"></span> e principalmente ao valor da variável <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P10\" title=\"3P10\">\n Observe que para <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1121.svg\" alt=\"$x=0$\" loading=\"lazy\"></span> a série se reduz a um único termo e, portanto, é uma série convergente (para <span class=\"MATH\"><img style=\"height: 1.52ex; vertical-align: -0.46ex; \" src=\"img/img1122.svg\" alt=\"$a_{0}$\" loading=\"lazy\"></span>). O que\n realmente interessa é se existem outros valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>, além de <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1121.svg\" alt=\"$x=0$\" loading=\"lazy\"></span>, para os quais a série de potências é\n convergente. Nestes termos um fato importante é a determinação dos valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> que tornam uma série de potências\n convergente, isto é, determinar os valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> para os quais a soma infinita existe.\n </p>\n <p class=\" unidade\" id=\"3P11\" title=\"3P11\">\n O conjunto dos valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> que tornam a série convergente é um intervalo, centrado em <span class=\"MATH\">0</span> e com raio <span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img1123.svg\" alt=\"$r>0$\" loading=\"lazy\"></span>. É um\n intervalo do tipo <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1124.svg\" alt=\"$(-r,r)$\" loading=\"lazy\"></span>, podendo ainda ser fechado em algum dos extremos. Para determinar este intervalo <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1124.svg\" alt=\"$(-r,r)$\" loading=\"lazy\"></span>\n usamos, em geral, o chamado teste da razão (Critério de D'Alembert).\n </p>\n <div id=\"3Teo2\" title=\"3Teo2\" class=\"unidade\"><b>Teorema <span class=\"arabic\">3</span>.<span class=\"arabic\">2</span></b> (Teste da razão) \n <i>Dada uma série <!-- MATH\n $\\sum b_{n}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1125.svg\" alt=\"$\\sum b_{n}$\" loading=\"lazy\"></span>, então\n </i><table width=\"90%\">\n <tbody><tr><td align=\"right\" valign=\"top\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1126.svg\" alt=\"$i)$\" loading=\"lazy\"></span></td><td valign=\"top\"> Se <!-- MATH\n $\\lim\\limits_{n \\to \\infty} \\left| \\frac{b_{n+1}}{b_{n}} \\right| = L < 1$\n -->\n <span class=\"MATH\"><img style=\"height: 4.12ex; vertical-align: -1.60ex; \" src=\"img/img1127.svg\" alt=\"$\\lim\\limits_{n \\to \\infty} \\left\\vert \\frac{b_{n+1}}{b_{n}} \\right\\vert = L < 1$\" loading=\"lazy\"></span>, a série é absolutamente\n convergente (e, portanto, convergente).\n </td></tr>\n <tr><td align=\"right\" valign=\"top\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1128.svg\" alt=\"$ii)$\" loading=\"lazy\"></span></td><td valign=\"top\"> Se <!-- MATH\n $\\lim\\limits_{n \\to \\infty} \\left| \\frac{b_{n+1}}{b_{n}} \\right| = L > 1$\n -->\n <span class=\"MATH\"><img style=\"height: 4.12ex; vertical-align: -1.60ex; \" src=\"img/img1129.svg\" alt=\"$\\lim\\limits_{n \\to \\infty} \\left\\vert \\frac{b_{n+1}}{b_{n}} \\right\\vert = L > 1$\" loading=\"lazy\"></span>, a série é divergente.\n </td></tr></tbody></table></div>\n \n <p class=\" unidade\" id=\"3P12\" title=\"3P12\">\n Para os valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> que tornam a série convergente, definimos uma função <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img299.svg\" alt=\"$f(x)$\" loading=\"lazy\"></span>, cujo domínio é o intervalo de\n convergência da série. O recíproco disto é uma pergunta mais interessante. Dada uma função <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img299.svg\" alt=\"$f(x)$\" loading=\"lazy\"></span> definida em algum\n intervalo <!-- MATH\n $I = (-r,r)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1130.svg\" alt=\"$I = (-r,r)$\" loading=\"lazy\"></span>, é possível obter uma série de potências <!-- MATH\n $\\sum a_{n}x^{n}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1131.svg\" alt=\"$\\sum a_{n}x^{n}$\" loading=\"lazy\"></span> de forma que <!-- MATH\n $f(x) = \\sum a_{n}x^{n}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1132.svg\" alt=\"$f(x) = \\sum a_{n}x^{n}$\" loading=\"lazy\"></span>\n para todo <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1133.svg\" alt=\"$x \\in I$\" loading=\"lazy\"></span>? Mais ainda, se existir tal série, como devem ser os coeficientes <span class=\"MATH\"><img style=\"height: 1.48ex; vertical-align: -0.41ex; \" src=\"img/img1117.svg\" alt=\"$a_{n}$\" loading=\"lazy\"></span>? A segunda pergunta\n é respondida pelo teorema de Maclaurin.\n \n </p>\n <div id=\"3Teo3\" title=\"3Teo3\" class=\"unidade\"><b>Teorema <span class=\"arabic\">3</span>.<span class=\"arabic\">3</span></b> (Teorema de Maclaurin) \n <i>Se <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img299.svg\" alt=\"$f(x)$\" loading=\"lazy\"></span> é uma função que admite uma representação por série de potências de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>,\n </i><!-- MATH\n \\begin{displaymath}\n f(x) = \\sum_{n=0}^{\\infty} a_{n} x^{n}\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P13\" title=\"3P13\">\n <img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1134.svg\" alt=\"$\\displaystyle f(x) = \\sum_{n=0}^{\\infty} a_{n} x^{n} $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P14\" style=\"text-indent: 0 !important;\" title=\"3P14\"><i>\n para todo <!-- MATH\n $x \\in (-r,r)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1135.svg\" alt=\"$x \\in (-r,r)$\" loading=\"lazy\"></span>, então <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img277.svg\" alt=\"$f$\" loading=\"lazy\"></span> é uma função infinitamente diferenciável no ponto <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1121.svg\" alt=\"$x=0$\" loading=\"lazy\"></span> e mais ainda,\n </i><!-- MATH\n \\begin{displaymath}\n f(x) = f(0) + f'(0)x + \\frac{f''(0)}{2} x^{2} + \\frac{f'''(0)}{3!} x^{3} + \\frac{f^{(4)}(0)}{4!} x^{4} + \\cdots + \\frac{f^{(n)}(0)}{n!} x^{n} + \\cdots\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P15\" title=\"3P15\">\n <img style=\"height: 5.01ex; vertical-align: -1.57ex; \" src=\"img/img1136.svg\" alt=\"$\\displaystyle f(x) = f(0) + f'(0)x + \\frac{f''(0)}{2} x^{2} + \\frac{f'''(0)}{3!...\n ... + \\frac{f^{(4)}(0)}{4!} x^{4} + \\cdots + \\frac{f^{(n)}(0)}{n!} x^{n} + \\cdots $\" loading=\"lazy\">\n </div><i>\n para todo <!-- MATH\n $x \\in (-r,r)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1135.svg\" alt=\"$x \\in (-r,r)$\" loading=\"lazy\"></span>.\n </i></div>\n \n <p class=\" unidade\" id=\"3P16\" title=\"3P16\">\n Este teorema nos diz principalmente que os coeficientes <span class=\"MATH\"><img style=\"height: 1.48ex; vertical-align: -0.41ex; \" src=\"img/img1117.svg\" alt=\"$a_{n}$\" loading=\"lazy\"></span>, da série de potências de uma função, são\n respectivamente <!-- MATH\n $\\frac{f^{(n)}(0)}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.36ex; vertical-align: -0.83ex; \" src=\"img/img1137.svg\" alt=\"$\\frac{f^{(n)}(0)}{n!}$\" loading=\"lazy\"></span>, sendo que a notação <span class=\"MATH\"><img style=\"height: 2.11ex; vertical-align: -0.10ex; \" src=\"img/img1138.svg\" alt=\"$f^{(n)}$\" loading=\"lazy\"></span> refere-se à derivada de ordem <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> da função <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img277.svg\" alt=\"$f$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P17\" title=\"3P17\">\n Para exemplificar o processo, vamos obter as séries de potência de algumas funções de interesse como <!-- MATH\n ${\\mathrm{senh}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1139.svg\" alt=\"${\\mathrm{senh}}x$\" loading=\"lazy\"></span>, <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1140.svg\" alt=\"$\\cosh\n x$\" loading=\"lazy\"></span>, <!-- MATH\n ${\\mathrm {sen}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1141.svg\" alt=\"${\\mathrm {sen}}x$\" loading=\"lazy\"></span>, <span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1142.svg\" alt=\"$\\cos x$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1143.svg\" alt=\"$e^{x}$\" loading=\"lazy\"></span>, assumindo que estas funções admitem uma representação em série de potências em algum\n intervalo <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1124.svg\" alt=\"$(-r,r)$\" loading=\"lazy\"></span>. Este não deve ser um trabalho difícil neste momento pois conhecemos as derivadas destas funções.\n Assim, parece não haver problemas significativos para a determinação dos coeficientes <span class=\"MATH\"><img style=\"height: 1.48ex; vertical-align: -0.41ex; \" src=\"img/img1117.svg\" alt=\"$a_{n}$\" loading=\"lazy\"></span> das séries de potências\n destas funções.\n </p>\n <p class=\" unidade\" id=\"3P18\" title=\"3P18\">\n Primeiramente, vamos à série de potências da função <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1143.svg\" alt=\"$e^{x}$\" loading=\"lazy\"></span>. Esperamos encontrar uma série de potências em <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>, de\n forma que,\n <!-- MATH\n \\begin{displaymath}\n e^{x} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + a_{5}x^{5} + \\cdots.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P19\" title=\"3P19\">\n <img style=\"height: 2.49ex; vertical-align: -0.46ex; \" src=\"img/img1144.svg\" alt=\"$\\displaystyle e^{x} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + a_{5}x^{5} + \\cdots. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P20\" title=\"3P20\">\n De acordo com o teorema de Maclaurin, devemos ter <!-- MATH\n $a_{n} = \\frac{f^{(n)}(0)}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.36ex; vertical-align: -0.83ex; \" src=\"img/img1145.svg\" alt=\"$a_{n} = \\frac{f^{(n)}(0)}{n!}$\" loading=\"lazy\"></span> para todo <!-- MATH\n $n \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1146.svg\" alt=\"$n \\in \\mathbb{N}$\" loading=\"lazy\"></span>. Como sabemos,\n a função exponencial <!-- MATH\n $f(x) = e^{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1147.svg\" alt=\"$f(x) = e^{x}$\" loading=\"lazy\"></span> possui derivadas de qualquer ordem contínuas e mais ainda <!-- MATH\n $f^{(n)}(x) = e^{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1148.svg\" alt=\"$f^{(n)}(x) = e^{x}$\" loading=\"lazy\"></span> para\n qualquer <!-- MATH\n $n \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1146.svg\" alt=\"$n \\in \\mathbb{N}$\" loading=\"lazy\"></span>. Então os coeficientes da série de Maclaurin ficam\n <!-- MATH\n \\begin{displaymath}\n a_{n} = \\frac{f^{(n)}(0)}{n!} = \\frac{e^{0}}{n!} = \\frac{1}{n!},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P21\" title=\"3P21\">\n <img style=\"height: 4.98ex; vertical-align: -1.55ex; \" src=\"img/img1149.svg\" alt=\"$\\displaystyle a_{n} = \\frac{f^{(n)}(0)}{n!} = \\frac{e^{0}}{n!} = \\frac{1}{n!}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P22\" style=\"text-indent: 0 !important;\" title=\"3P22\">\n e assim temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P23\" title=\"3P23\"><a id=\"exserie\"></a><!-- MATH\n \\begin{equation}\n e^{x} = 1 + x + \\frac{1}{2}x^{2} + \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} + \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} + \\frac{1}{7!}x^{7} + \\cdots.\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1150.svg\" alt=\"$\\displaystyle e^{x} = 1 + x + \\frac{1}{2}x^{2} + \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} + \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} + \\frac{1}{7!}x^{7} + \\cdots.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">3</span>) </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P24\" title=\"3P24\">\n Podemos determinar (pelo teste da razão) que a série do lado direito converge para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, já que para qualquer <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>,\n <!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left| \\frac{\\frac{1}{(n+1)!} x^{n+1}}{ \\frac{1}{n!} x^{n}} \\right|\n = \\lim_{n \\to \\infty} \\left| \\frac{n!}{(n+1)!} \\frac{x^{n+1}}{x^{n}} \\right| = \\lim_{n \\to \\infty} \\frac{1}{n+1} |x| = 0 < 1, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P25\" title=\"3P25\">\n <img style=\"height: 6.68ex; vertical-align: -2.74ex; \" src=\"img/img1152.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert \\frac{\\frac{1}{(n+1)!} x^{n+1}}{ \\...\n ...}{x^{n}} \\right\\vert = \\lim_{n \\to \\infty} \\frac{1}{n+1} \\vert x\\vert = 0 < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P26\" style=\"text-indent: 0 !important;\" title=\"3P26\">\n e, portanto, a igualdade (<a href=\"#exserie\">3.3</a>) é válida para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P27\" title=\"3P27\">\n Consideremos agora a função <!-- MATH\n $f(x) = {\\mathrm{senh}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1153.svg\" alt=\"$f(x) = {\\mathrm{senh}}x$\" loading=\"lazy\"></span>. Queremos determinar os coeficientes <!-- MATH\n $a_{n} = \\frac{f^{(n)}(0)}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.36ex; vertical-align: -0.83ex; \" src=\"img/img1145.svg\" alt=\"$a_{n} = \\frac{f^{(n)}(0)}{n!}$\" loading=\"lazy\"></span> para\n todo <!-- MATH\n $n \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1146.svg\" alt=\"$n \\in \\mathbb{N}$\" loading=\"lazy\"></span>. Dos resultados dos capítulos anteriores, temos que se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é par, <!-- MATH\n $f^{(n)}(x) = {\\mathrm{senh}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1154.svg\" alt=\"$f^{(n)}(x) = {\\mathrm{senh}}x$\" loading=\"lazy\"></span> e se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é\n ímpar <!-- MATH\n $f^{(n)}(x) = \\cosh x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1155.svg\" alt=\"$f^{(n)}(x) = \\cosh x$\" loading=\"lazy\"></span>. Desta forma, os coeficientes são dados por\n <!-- MATH\n \\begin{displaymath}\n a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{{\\mathrm{senh}}0}{n!} = \\dfrac{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é par}, \\\\\n & \\\\\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{\\cosh 0}{n!} = \\dfrac{1}{n!} & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P28\" title=\"3P28\">\n <img style=\"height: 12.90ex; vertical-align: -5.94ex; \" src=\"img/img1156.svg\" alt=\"$\\displaystyle a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfr...\n ...dfrac{1}{n!} & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P29\" title=\"3P29\">\n Substituindo estes coeficientes na série de potências temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P30\" title=\"3P30\"><a id=\"sinhxserie\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm{senh}}x = x + \\frac{1}{3!}x^{3} + \\frac{1}{5!}x^{5} + \\frac{1}{7!}x^{7} + \\frac{1}{9!}x^{9} + \\frac{1}{11!}x^{11} + \\cdots.\n \\end{equation}\n -->\n <table >\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1157.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}x = x + \\frac{1}{3!}x^{3} + \\frac{1}{5!}x^{5} + \\frac{1}{7!}x^{7} + \\frac{1}{9!}x^{9} + \\frac{1}{11!}x^{11} + \\cdots.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">4</span>)</td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P31\" title=\"3P31\">\n A série do lado direito converge (pelo teste da razão) para qualquer <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, pois para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span> temos\n <!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left| \\frac{\\frac{1}{(2n+1)!} x^{2n+1}}{ \\frac{1}{(2n-1)!} x^{2n-1}} \\right|\n = \\lim_{n \\to \\infty} \\left| \\frac{(2n-1)!}{(2n+1)!} \\frac{x^{2n+1}}{x^{2n-1}} \\right| = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} |x^{2}| = 0 < 1, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P32\" title=\"3P32\">\n <img style=\"height: 6.84ex; vertical-align: -2.91ex; \" src=\"img/img1158.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert \\frac{\\frac{1}{(2n+1)!} x^{2n+1}}{...\n ...\\right\\vert = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} \\vert x^{2}\\vert = 0 < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P33\" style=\"text-indent: 0 !important;\" title=\"3P33\">\n e, portanto, a igualdade (<a href=\"#sinhxserie\">3.4</a>) é válida para todo <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> real.\n </p>\n <p class=\" unidade\" id=\"3P34\" title=\"3P34\">\n Com raciocínio similar desenvolvemos a série de potências para a função <!-- MATH\n $f(x) = \\cosh x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1159.svg\" alt=\"$f(x) = \\cosh x$\" loading=\"lazy\"></span>. Lembremos que agora,\n <!-- MATH\n $f^{(n)}(x) = \\cosh x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1155.svg\" alt=\"$f^{(n)}(x) = \\cosh x$\" loading=\"lazy\"></span> se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é par e <!-- MATH\n $f^{(n)}(x) = {\\mathrm{senh}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1154.svg\" alt=\"$f^{(n)}(x) = {\\mathrm{senh}}x$\" loading=\"lazy\"></span> se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é ímpar. Então, contrariamente ao caso anterior,\n <!-- MATH\n \\begin{displaymath}\n a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{\\cosh 0}{n!} = \\dfrac{1}{n!} & \\text{se} \\quad n \\quad \\text{é par}, \\\\\n & \\\\\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{{\\mathrm{senh}}0}{n!} = \\dfrac{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P35\" title=\"3P35\">\n <img style=\"height: 12.90ex; vertical-align: -5.94ex; \" src=\"img/img1160.svg\" alt=\"$\\displaystyle a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfr...\n ...c{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P36\" title=\"3P36\">\n Desta forma, temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P37\" title=\"3P37\"><a id=\"coshxserie\"></a><!-- MATH\n \\begin{equation}\n \\cosh x = 1 + \\frac{1}{2!}x^{2} + \\frac{1}{4!}x^{4} + \\frac{1}{6!}x^{6} + \\frac{1}{8!}x^{8} + \\frac{1}{10!}x^{10} + \\cdots,\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1161.svg\" alt=\"$\\displaystyle \\cosh x = 1 + \\frac{1}{2!}x^{2} + \\frac{1}{4!}x^{4} + \\frac{1}{6!}x^{6} + \\frac{1}{8!}x^{8} + \\frac{1}{10!}x^{10} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">5</span>) </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P38\" style=\"text-indent: 0 !important;\" title=\"3P38\">\n sendo também esta igualdade verdadeira para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P39\" title=\"3P39\">\n Dada agora a função <!-- MATH\n $f(x) = {\\mathrm {sen}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1162.svg\" alt=\"$f(x) = {\\mathrm {sen}}x$\" loading=\"lazy\"></span>, queremos determinar para todo <!-- MATH\n $n \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1146.svg\" alt=\"$n \\in \\mathbb{N}$\" loading=\"lazy\"></span> os coeficientes <!-- MATH\n $a_{n} =\n \\frac{f^{(n)}(0)}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.36ex; vertical-align: -0.83ex; \" src=\"img/img1145.svg\" alt=\"$a_{n} = \\frac{f^{(n)}(0)}{n!}$\" loading=\"lazy\"></span> da série de potências de <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img299.svg\" alt=\"$f(x)$\" loading=\"lazy\"></span>. Dos resultados anteriores, sabemos que se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é par,\n <!-- MATH\n $f^{(n)}(x) = (-1)^{\\frac{n}{2}} {\\mathrm {sen}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.71ex; vertical-align: -0.62ex; \" src=\"img/img1163.svg\" alt=\"$f^{(n)}(x) = (-1)^{\\frac{n}{2}} {\\mathrm {sen}}x$\" loading=\"lazy\"></span> e se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é ímpar <!-- MATH\n $f^{(n)}(x) = (-1)^{\\frac{n-1}{2}} \\cos x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.98ex; vertical-align: -0.62ex; \" src=\"img/img1164.svg\" alt=\"$f^{(n)}(x) = (-1)^{\\frac{n-1}{2}} \\cos x$\" loading=\"lazy\"></span>. Desta forma, os\n coeficientes são\n <!-- MATH\n \\begin{displaymath}\n a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{(-1)^{\\frac{n}{2}} {\\mathrm {sen}}0}{n!} = \\dfrac{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é par}, \\\\\n & \\\\\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{(-1)^{\\frac{n-1}{2}} \\cos 0}{n!} = \\dfrac{(-1)^{\\frac{n-1}{2}}}{n!} & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P40\" title=\"3P40\">\n <img style=\"height: 13.31ex; vertical-align: -6.15ex; \" src=\"img/img1165.svg\" alt=\"$\\displaystyle a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfr...\n ...-1}{2}}}{n!} & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P41\" title=\"3P41\">\n Observe que os termos <!-- MATH\n $(-1)^{\\frac{n-1}{2}}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.98ex; vertical-align: -0.62ex; \" src=\"img/img1166.svg\" alt=\"$(-1)^{\\frac{n-1}{2}}$\" loading=\"lazy\"></span> para <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> ímpar, são alternadamente 1 e -1. Substituindo estes coeficientes\n na série de potências temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P42\" title=\"3P42\"><a id=\"sinxserie\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm {sen}}x = x - \\frac{1}{3!}x^{3} + \\frac{1}{5!}x^{5} - \\frac{1}{7!}x^{7} + \\frac{1}{9!}x^{9} - \\frac{1}{11!}x^{11} + \\cdots.\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1167.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}x = x - \\frac{1}{3!}x^{3} + \\frac{1}{5!}x^{5} - \\frac{1}{7!}x^{7} + \\frac{1}{9!}x^{9} - \\frac{1}{11!}x^{11} + \\cdots.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">6</span>) </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P43\" title=\"3P43\">\n A série do lado direito converge (pelo teste da razão) para qualquer <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, já que\n <!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left|- \\frac{\\frac{1}{(2n+1)!} x^{2n+1}}{ \\frac{1}{(2n-1)!} x^{2n-1}} \\right|\n = \\lim_{n \\to \\infty} \\left| \\frac{(2n-1)!}{(2n+1)!} \\frac{x^{2n+1}}{x^{2n-1}} \\right| = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} |x^{2}| = 0 < 1, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P44\" title=\"3P44\">\n <img style=\"height: 6.84ex; vertical-align: -2.91ex; \" src=\"img/img1168.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert- \\frac{\\frac{1}{(2n+1)!} x^{2n+1}}...\n ...\\right\\vert = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} \\vert x^{2}\\vert = 0 < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P45\" style=\"text-indent: 0 !important;\" title=\"3P45\">\n para qualquer <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Segue que a igualdade (<a href=\"#sinxserie\">3.6</a>) é válida para todo <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> real.\n </p>\n <p class=\" unidade\" id=\"3P46\" title=\"3P46\">\n Com raciocínio similar a este abordamos a função <!-- MATH\n $f(x) = \\cos x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1169.svg\" alt=\"$f(x) = \\cos x$\" loading=\"lazy\"></span>. Lembremos que, <!-- MATH\n $f^{(n)}(x) = (-1)^{\\frac{n}{2}} \\cos\n x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.71ex; vertical-align: -0.62ex; \" src=\"img/img1170.svg\" alt=\"$f^{(n)}(x) = (-1)^{\\frac{n}{2}} \\cos\n x$\" loading=\"lazy\"></span> se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é par e <!-- MATH\n $f^{(n)}(x) = (-1)^{\\frac{n+1}{2}} {\\mathrm {sen}}x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.98ex; vertical-align: -0.62ex; \" src=\"img/img1171.svg\" alt=\"$f^{(n)}(x) = (-1)^{\\frac{n+1}{2}} {\\mathrm {sen}}x$\" loading=\"lazy\"></span> se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> é ímpar. Então, contrariamente ao caso anterior,\n <!-- MATH\n \\begin{displaymath}\n a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{(-1)^{\\frac{n}{2}} \\cos 0}{n!} = \\dfrac{(-1)^{\\frac{n}{2}}}{n!} & \\text{se} \\quad n \\quad \\text{é par}, \\\\\n & \\\\\n \\dfrac{f^{(n)}(0)}{n!} = \\dfrac{(-1)^{\\frac{n+1}{2}} {\\mathrm {sen}}0}{n!} = \\dfrac{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P47\" title=\"3P47\">\n <img style=\"height: 13.31ex; vertical-align: -6.15ex; \" src=\"img/img1172.svg\" alt=\"$\\displaystyle a_{n} = \\left\\{ \\begin{array}{ll}\n \\dfrac{f^{(n)}(0)}{n!} = \\dfr...\n ...c{0}{n!} = 0 & \\text{se} \\quad n \\quad \\text{é ímpar}.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P48\" title=\"3P48\">\n Desta forma, temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P49\" title=\"3P49\"><a id=\"cosxserie\"></a><!-- MATH\n \\begin{equation}\n \\cos x = 1 - \\frac{1}{2!}x^{2} + \\frac{1}{4!}x^{4} - \\frac{1}{6!}x^{6} + \\frac{1}{8!}x^{8} - \\frac{1}{10!}x^{10} + \\cdots,\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1173.svg\" alt=\"$\\displaystyle \\cos x = 1 - \\frac{1}{2!}x^{2} + \\frac{1}{4!}x^{4} - \\frac{1}{6!}x^{6} + \\frac{1}{8!}x^{8} - \\frac{1}{10!}x^{10} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">7</span>)</td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P50\" style=\"text-indent: 0 !important;\" title=\"3P50\">\n sendo também esta igualdade verdadeira para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P51\" title=\"3P51\">\n Vamos usar agora as séries de potência obtidas anteriormente para verificar a validade das igualdades em\n (<a href=\"#idexpo\">3.1</a>). Consideremos primeiro a série de potência da função exponencial em <a href=\"#secspot\">3.1</a>,\n <!-- MATH\n \\begin{displaymath}\n e^{x} = 1 + x + \\frac{1}{2}x^{2} + \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} + \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} + \\frac{1}{7!}x^{7} + \\cdots,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P52\" title=\"3P52\">\n <img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1174.svg\" alt=\"$\\displaystyle e^{x} = 1 + x + \\frac{1}{2}x^{2} + \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} + \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} + \\frac{1}{7!}x^{7} + \\cdots, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P53\" title=\"3P53\">\n válida para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Naturalmente se substituirmos na série <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> por <span class=\"MATH\"><img style=\"height: 1.59ex; vertical-align: -0.27ex; \" src=\"img/img1175.svg\" alt=\"$-x$\" loading=\"lazy\"></span> obtemos a série de potências para a\n função <span class=\"MATH\"><img style=\"height: 1.85ex; vertical-align: -0.10ex; \" src=\"img/img1176.svg\" alt=\"$e^{-x}$\" loading=\"lazy\"></span>. Isto também pode ser feito como anteriormente, determinando-se os coeficientes <!-- MATH\n $a_{n} =\n \\frac{f^{(n)}(0)}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.36ex; vertical-align: -0.83ex; \" src=\"img/img1145.svg\" alt=\"$a_{n} = \\frac{f^{(n)}(0)}{n!}$\" loading=\"lazy\"></span> para a série de potências da função <!-- MATH\n $f(x) = e^{-x}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.62ex; \" src=\"img/img1177.svg\" alt=\"$f(x) = e^{-x}$\" loading=\"lazy\"></span>. Isto nos leva aos coeficientes <!-- MATH\n $a_{n} =\n \\frac{(-1)^{n}}{n!}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.11ex; vertical-align: -0.83ex; \" src=\"img/img1178.svg\" alt=\"$a_{n} =\n \\frac{(-1)^{n}}{n!}$\" loading=\"lazy\"></span>. De qualquer forma teremos\n <!-- MATH\n \\begin{displaymath}\n e^{-x} = 1 - x + \\frac{1}{2}x^{2} - \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} - \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} - \\frac{1}{7!}x^{7} + \\cdots,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P54\" title=\"3P54\">\n <img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1179.svg\" alt=\"$\\displaystyle e^{-x} = 1 - x + \\frac{1}{2}x^{2} - \\frac{1}{3!}x^{3} + \\frac{1}{4!}x^{4} - \\frac{1}{5!}x^{5} + \\frac{1}{6!}x^{6} - \\frac{1}{7!}x^{7} + \\cdots, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P55\" style=\"text-indent: 0 !important;\" title=\"3P55\">\n para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Somando as série de potências de <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1143.svg\" alt=\"$e^{x}$\" loading=\"lazy\"></span> e de <span class=\"MATH\"><img style=\"height: 1.85ex; vertical-align: -0.10ex; \" src=\"img/img1176.svg\" alt=\"$e^{-x}$\" loading=\"lazy\"></span> obtemos,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P56\" title=\"3P56\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1180.svg\" alt=\"$\\displaystyle e^{x} + e^{-x}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1181.svg\" alt=\"$\\displaystyle = 2 + 2\\frac{1}{2}x^{2} + 2\\frac{1}{4!}x^{4} + 2\\frac{1}{6!}x^{6} + 2\\frac{1}{8!}x^{8} + 2\\frac{1}{10!}x^{10} + \\cdots$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.24ex; vertical-align: -2.10ex; \" src=\"img/img1182.svg\" alt=\"$\\displaystyle = 2 \\left( 1 + \\frac{1}{2}x^{2} + \\frac{1}{4!}x^{4} + \\frac{1}{6!}x^{6} + \\frac{1}{8!}x^{8} + \\frac{1}{10!}x^{10} + \\cdots \\right).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P57\" title=\"3P57\">\n Agora lembremos que a série do lado direito é exatamente a série de potências da função cosseno hiperbólico (Ver\n (<a href=\"#coshxserie\">3.5</a>)). Desta forma temos que\n <!-- MATH\n \\begin{displaymath}\n e^{x} + e^{-x} = 2 \\cosh x,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P58\" title=\"3P58\">\n <img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1183.svg\" alt=\"$\\displaystyle e^{x} + e^{-x} = 2 \\cosh x, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P59\" style=\"text-indent: 0 !important;\" title=\"3P59\">\n donde segue que\n <!-- MATH\n \\begin{displaymath}\n \\cosh x = \\frac{e^{x} + e^{-x}}{2}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P60\" title=\"3P60\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1184.svg\" alt=\"$\\displaystyle \\cosh x = \\frac{e^{x} + e^{-x}}{2}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P61\" title=\"3P61\">\n Por outro lado, fazendo a diferença entre as séries de potências das funções <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1143.svg\" alt=\"$e^{x}$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.85ex; vertical-align: -0.10ex; \" src=\"img/img1176.svg\" alt=\"$e^{-x}$\" loading=\"lazy\"></span>, temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P62\" title=\"3P62\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1185.svg\" alt=\"$\\displaystyle e^{x} - e^{-x}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1186.svg\" alt=\"$\\displaystyle = 2x + 2\\frac{1}{3!}x^{3} + 2\\frac{1}{5!}x^{5} + 2\\frac{1}{7!}x^{7} + 2\\frac{1}{9!}x^{9} + 2\\frac{1}{11!}x^{11} + \\cdots$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.24ex; vertical-align: -2.10ex; \" src=\"img/img1187.svg\" alt=\"$\\displaystyle = 2 \\left( x + \\frac{1}{3!}x^{3} + \\frac{1}{5!}x^{5} + \\frac{1}{7!}x^{7} + \\frac{1}{9!}x^{9} + \\frac{1}{11!}x^{11} + \\cdots \\right),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P63\" style=\"text-indent: 0 !important;\" title=\"3P63\">\n e lembrando que o lado direito é a série de potências da função seno hiperbólico (Ver (<a href=\"#sinhxserie\">3.4</a>)), temos que\n <!-- MATH\n \\begin{displaymath}\n e^{x} - e^{-x} = 2 {\\mathrm{senh}}x,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P64\" title=\"3P64\">\n <img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1188.svg\" alt=\"$\\displaystyle e^{x} - e^{-x} = 2 {\\mathrm{senh}}x, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P65\" style=\"text-indent: 0 !important;\" title=\"3P65\">\n donde segue\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{senh}}x = \\frac{e^{x} - e^{-x}}{2}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P66\" title=\"3P66\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1189.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}x = \\frac{e^{x} - e^{-x}}{2}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P67\" title=\"3P67\">\n Fica assim verificada a validade das fórmulas exponenciais, que são comumente utilizadas para definir as funções\n trigonométricas hiperbólicas. Com estas duas igualdades, podemos escrever as demais funções trigonométricas\n hiperbólicas também em termos da função exponencial. São\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P68\" title=\"3P68\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1190.svg\" alt=\"$\\displaystyle {\\mathrm {tgh}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.90ex; vertical-align: -1.72ex; \" src=\"img/img1191.svg\" alt=\"$\\displaystyle = \\frac{{\\mathrm{senh}}x}{\\cosh x} = \\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}},$\" loading=\"lazy\"> para<img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1192.svg\" alt=\"$\\displaystyle \\quad x \\in \\mathbb{R},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1193.svg\" alt=\"$\\displaystyle {\\mathrm{ctgh}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.90ex; vertical-align: -1.72ex; \" src=\"img/img1194.svg\" alt=\"$\\displaystyle = \\frac{\\cosh x}{{\\mathrm{senh}}x} = \\frac{e^{x} + e^{-x}}{e^{x} - e^{-x}},$\" loading=\"lazy\"> para<img style=\"height: 2.06ex; vertical-align: -0.51ex; \" src=\"img/img1195.svg\" alt=\"$\\displaystyle \\quad x \\neq 0,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1196.svg\" alt=\"$\\displaystyle {\\mathrm{sech}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1197.svg\" alt=\"$\\displaystyle = \\frac{1}{\\cosh x} = \\frac{2}{e^{x} + e^{-x}},$\" loading=\"lazy\"> para<img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1192.svg\" alt=\"$\\displaystyle \\quad x \\in \\mathbb{R},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1198.svg\" alt=\"$\\displaystyle {\\mathrm{csch}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1199.svg\" alt=\"$\\displaystyle = \\frac{1}{{\\mathrm{senh}}x} = \\frac{2}{e^{x} - e^{-x}},$\" loading=\"lazy\"> para<img style=\"height: 2.06ex; vertical-align: -0.51ex; \" src=\"img/img1200.svg\" alt=\"$\\displaystyle \\quad x \\neq 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n\n:::\n\n## 3.2 Método das equações diferenciais {#SECTION00720000000000000000}\n\n::: {.raw_html}\n \n <p class=\" unidade\" id=\"3P69\" title=\"3P69\">\n Nesta seção, provaremos as identidades em (<a href=\"#idexpo\">3.1</a>) usando o método das equações diferenciais. Precisamos\n naturalmente alguns resultados em relação às equações diferenciais. Para um estudo mais aprofundado sobre equações\n diferenciais recomendamos [<a href=\"/trigonometria-hiperbolica/referencias#Zill\">10</a>, Zill].\n </p>\n \n <div><p style=\"text-indent: 0 !important;\"><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">4</span></b> \n Uma equação diferencial ordinária, de ordem <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span>, é uma equação que envolve uma variável real independente <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>, uma\n função <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1201.svg\" alt=\"$y = f(x)$\" loading=\"lazy\"></span> e suas derivadas <!-- MATH\n $y', y'', \\dots, y^{(n)}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1202.svg\" alt=\"$y', y'', \\dots, y^{(n)}$\" loading=\"lazy\"></span>, de forma que o coeficiente de <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1203.svg\" alt=\"$y^{(n)}$\" loading=\"lazy\"></span> seja não nulo.\n </p></div>\n \n <p class=\" unidade\" id=\"3P70\" title=\"3P70\">\n São exemplos de equações diferencias ordinárias:\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P71\" title=\"3P71\"><table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1204.svg\" alt=\"$\\displaystyle y'' + y = {\\mathrm {sen}}(2x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n <a id=\"EDO1\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">8</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.31ex; vertical-align: -0.50ex; \" src=\"img/img1205.svg\" alt=\"$\\displaystyle y \\cdot y' + x = 0$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n <a id=\"EDO2\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">9</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.61ex; vertical-align: -0.50ex; \" src=\"img/img1206.svg\" alt=\"$\\displaystyle y^{(4)} + xy'' +e^{x}y = 0$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1207.svg\" alt=\"$\\displaystyle (y')^{2} + 3xy' - 4x^{2} = 0$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n <a id=\"EDO4\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">10</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.53ex; vertical-align: -0.50ex; \" src=\"img/img1208.svg\" alt=\"$\\displaystyle x^{3}y''' +2x^{2}y'' - xy' + y = 12x^{2}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n <a id=\"EDO5\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">11</span>)</a></td></tr>\n </tbody></table></div>\n \n \n <div><p style=\"text-indent: 0 !important;\"><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">5</span></b> \n Uma equação diferencial ordinária linear, de ordem <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span>, é uma equação diferencial que seja linear nas componentes\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1209.svg\" alt=\"$y^{(k)}$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $k = 0, 1, \\dots, n$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1210.svg\" alt=\"$k = 0, 1, \\dots, n$\" loading=\"lazy\"></span>. É uma expressão da forma,\n <!-- MATH\n \\begin{displaymath}\n a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \\cdots + a_{2}(x) y^{''} + a_{1}(x) y^{'} + a_{0}(x) y = g(x),\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P72\" title=\"3P72\">\n <img style=\"height: 2.77ex; vertical-align: -0.62ex; \" src=\"img/img1211.svg\" alt=\"$\\displaystyle a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \\cdots + a_{2}(x) y^{''} + a_{1}(x) y^{'} + a_{0}(x) y = g(x), $\" loading=\"lazy\">\n </div>\n sendo que as funções coeficientes <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1212.svg\" alt=\"$a_{k}(x)$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $k = 0,1, \\dots, n$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1210.svg\" alt=\"$k = 0, 1, \\dots, n$\" loading=\"lazy\"></span> e a função <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img300.svg\" alt=\"$g(x)$\" loading=\"lazy\"></span>, são contínuas para todo\n <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> em um certo intervalo de interesse <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>. Também <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1213.svg\" alt=\"$a_{n}(x)$\" loading=\"lazy\"></span> é não identicamente nula em <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>. Quando <!-- MATH\n $g(x) \\equiv 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1214.svg\" alt=\"$g(x) \\equiv 0$\" loading=\"lazy\"></span>,\n então dizemos que a equação diferencial é homogênea.\n </p></div>\n \n <div><p style=\"text-indent: 0 !important;\"><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">6</span></b> \n Qualquer função <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1201.svg\" alt=\"$y = f(x)$\" loading=\"lazy\"></span>, definida num intervalo <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>, que satisfaz a equação diferencial neste intervalo, é dita uma\n solução para a equação diferencial em <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>.\n </p></div>\n \n <p class=\" unidade\" id=\"3P73\" title=\"3P73\">\n As funções <!-- MATH\n $y = -\\frac{1}{3}{\\mathrm {sen}}(2x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.83ex; vertical-align: -0.85ex; \" src=\"img/img1215.svg\" alt=\"$y = -\\frac{1}{3}{\\mathrm {sen}}(2x)$\" loading=\"lazy\"></span>, <!-- MATH\n $y = -2x^{2}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.42ex; vertical-align: -0.50ex; \" src=\"img/img1216.svg\" alt=\"$y = -2x^{2}$\" loading=\"lazy\"></span> e <!-- MATH\n $y = x + x\\ln x + 4x^{2}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.42ex; vertical-align: -0.50ex; \" src=\"img/img1217.svg\" alt=\"$y = x + x\\ln x + 4x^{2}$\" loading=\"lazy\"></span> (<span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img637.svg\" alt=\"$x>0$\" loading=\"lazy\"></span>), são soluções das equações\n diferenciais (<a href=\"#EDO1\">3.8</a>), (<a href=\"#EDO4\">3.10</a>) e (<a href=\"#EDO5\">3.11</a>), respectivamente. A solução da equação diferencial (<a href=\"#EDO2\">3.9</a>)\n é dada implicitamente por <!-- MATH\n $x^{2} + y^{2} = 4$\n -->\n <span class=\"MATH\"><img style=\"height: 2.42ex; vertical-align: -0.50ex; \" src=\"img/img1218.svg\" alt=\"$x^{2} + y^{2} = 4$\" loading=\"lazy\"></span>, para <!-- MATH\n $-2 < x < 2$\n -->\n <span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1219.svg\" alt=\"$-2 < x < 2$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P74\" title=\"3P74\">\n Embora a ideia seja bastante simples, encontrar uma solução para uma equação diferencial dada, não é tarefa simples. Os\n métodos conhecidos nos permitem determinar soluções de uma classe muito pequena de equações diferenciais. Mesmo assim,\n algumas destas equações não possuem solução explícita.\n </p>\n <p class=\" unidade\" id=\"3P75\" title=\"3P75\">\n Não é do nosso interesse estudar aqui os métodos para obtenção de soluções de uma equação diferencial. Entretanto é\n importante saber que nem sempre uma solução para uma equação diferencial é única. Podemos verificar que a função\n <!-- MATH\n \\begin{displaymath}\n y = k_{1} \\cos(x) + k_{2} {\\mathrm {sen}}(x),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P76\" title=\"3P76\">\n <img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1220.svg\" alt=\"$\\displaystyle y = k_{1} \\cos(x) + k_{2} {\\mathrm {sen}}(x), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P77\" style=\"text-indent: 0 !important;\" title=\"3P77\">\n é uma solução da equação <!-- MATH\n $y'' + y = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.21ex; vertical-align: -0.50ex; \" src=\"img/img1221.svg\" alt=\"$y'' + y = 0$\" loading=\"lazy\"></span>, para quaisquer valores reais de <span class=\"MATH\"><img style=\"height: 2.00ex; vertical-align: -0.45ex; \" src=\"img/img1222.svg\" alt=\"$k_{1}$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.99ex; vertical-align: -0.44ex; \" src=\"img/img1223.svg\" alt=\"$k_{2}$\" loading=\"lazy\"></span>. Para podermos determinar os\n valores de <span class=\"MATH\"><img style=\"height: 2.00ex; vertical-align: -0.45ex; \" src=\"img/img1222.svg\" alt=\"$k_{1}$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.99ex; vertical-align: -0.44ex; \" src=\"img/img1223.svg\" alt=\"$k_{2}$\" loading=\"lazy\"></span> são necessárias informações adicionais, chamadas de condições iniciais. Como veremos mais\n tarde, dentro de certas hipóteses, uma equação diferencial munida de condições iniciais possui solução única.\n </p>\n \n <div><p style=\"text-indent: 0 !important;\"><b>Definição <span class=\"arabic\">3</span>..<span class=\"arabic\">7</span></b> \n Um problema de valor inicial, ou PVI, consiste de uma equação diferencial ordinária, de ordem <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span>, juntamente com <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span>\n restrições. Tais restrições são chamadas de condições iniciais. É um problema da forma,\n <!-- MATH\n \\begin{displaymath}\n \\left\\{ \\begin{array}{l}\n a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \\cdots + a_{2}(x) y'' + a_{1}(x) y' + a_{0}(x) y = g(x), \\\\\n y^{(n-1)}(x_{0}) = y_{n-1}, \\quad \\dots, \\quad y''(x_{0}) = y_{2}, \\quad y'(x_{0}) = y_{1}, \\quad y(x_{0}) = y_{0}\n \\end{array} \\right. \n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P78\" title=\"3P78\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1224.svg\" alt=\"$\\displaystyle \\left\\{ \\begin{array}{l}\n a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)}...\n ... = y_{2}, \\quad y'(x_{0}) = y_{1}, \\quad y(x_{0}) = y_{0}\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n sendo que <span class=\"MATH\"><img style=\"height: 1.52ex; vertical-align: -0.46ex; \" src=\"img/img1225.svg\" alt=\"$x_{0}$\" loading=\"lazy\"></span> é um ponto de interesse no intervalo <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span> e os valores <span class=\"MATH\"><img style=\"height: 1.56ex; vertical-align: -0.50ex; \" src=\"img/img1226.svg\" alt=\"$y_{k}$\" loading=\"lazy\"></span>, para <!-- MATH\n $k = 0,1, \\dots, n-1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1227.svg\" alt=\"$k = 0,1, \\dots, n-1$\" loading=\"lazy\"></span>, são\n números reais conhecidos.\n </p></div>\n \n <p class=\" unidade\" id=\"3P79\" title=\"3P79\">\n Cuidado para não confundir um PVI com uma equação diferencial. Um PVI é um conjunto de uma equação diferencial\n juntamente com condições iniciais.\n </p>\n <p class=\" unidade\" id=\"3P80\" title=\"3P80\">\n Como exemplo, vamos agora considerar a equação diferencial de ordem 2, mencionada anteriormente <!-- MATH\n $y'' + y = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.21ex; vertical-align: -0.50ex; \" src=\"img/img1221.svg\" alt=\"$y'' + y = 0$\" loading=\"lazy\"></span>, a sua\n “família” de soluções dada por <!-- MATH\n $y = k_{1} \\cos(x) + k_{2} {\\mathrm {sen}}(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1228.svg\" alt=\"$y = k_{1} \\cos(x) + k_{2} {\\mathrm {sen}}(x)$\" loading=\"lazy\"></span> e impor duas condições iniciais que permitirão\n determinar os valores de <span class=\"MATH\"><img style=\"height: 2.00ex; vertical-align: -0.45ex; \" src=\"img/img1222.svg\" alt=\"$k_{1}$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.99ex; vertical-align: -0.44ex; \" src=\"img/img1223.svg\" alt=\"$k_{2}$\" loading=\"lazy\"></span>. Tomemos o PVI,\n <!-- MATH\n \\begin{displaymath}\n \\left\\{\\begin{array}{l} y'' + y = 0 \\\\y'(0) = 1, \\quad y(0) = 2, \\end{array} \\right.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P81\" title=\"3P81\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1229.svg\" alt=\"$\\displaystyle \\left\\{\\begin{array}{l} y'' + y = 0 \\\\ y'(0) = 1, \\quad y(0) = 2, \\end{array} \\right. $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P82\" style=\"text-indent: 0 !important;\" title=\"3P82\">\n e substituindo as duas condições iniciais, temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P83\" title=\"3P83\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.11ex; \" src=\"img/img1230.svg\" alt=\"$\\displaystyle 1$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1231.svg\" alt=\"$\\displaystyle = y'(0) = - k_{1} {\\mathrm {sen}}(0) + k_{2} \\cos(0) = k_{2},$\" loading=\"lazy\"> e</span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1232.svg\" alt=\"$\\displaystyle 2$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1233.svg\" alt=\"$\\displaystyle = y(0) = k_{1} \\cos(0) + k_{2} {\\mathrm {sen}}(0) = k_{1},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P84\" style=\"text-indent: 0 !important;\" title=\"3P84\">\n o que nos leva a uma única solução do PVI dado, que é\n <!-- MATH\n \\begin{displaymath}\n y = 2 \\cos(x) + {\\mathrm {sen}}(x).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P85\" title=\"3P85\">\n <img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1234.svg\" alt=\"$\\displaystyle y = 2 \\cos(x) + {\\mathrm {sen}}(x). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P86\" title=\"3P86\">\n O próximo teorema, é a chave para o nosso objetivo. Sua demonstração é que não é do nosso interesse, pois além de não\n ser o objetivo principal deste capítulo, é um tanto complexa e exige ferramentas que não abordamos como por exemplo o\n teorema de ponto fixo de Banach. Desta forma, vamos omitir a sua demonstração. O leitor interessado nesta demonstração\n poderá consultar algum texto de Equações Diferenciais (ordinárias). Sugerimos [<a href=\"/trigonometria-hiperbolica/referencias#Zill\">10</a>, Zill].\n </p>\n \n <div id=\"3Teo8\" title=\"3Teo8\" class=\"unidade\"><a id=\"teoPicard\"><b>Teorema <span class=\"arabic\">3</span>.<span class=\"arabic\">8</span></b></a> (Teorema de Picard) \n <i>Se as funções <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1212.svg\" alt=\"$a_{k}(x)$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $k = 0,1,\\dots, n$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1210.svg\" alt=\"$k = 0, 1, \\dots, n$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img300.svg\" alt=\"$g(x)$\" loading=\"lazy\"></span>, forem contínuas em um intervalo <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>, com <!-- MATH\n $a_{n}(x)\n \\neq 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1235.svg\" alt=\"$a_{n}(x)\n \\neq 0$\" loading=\"lazy\"></span> para todo <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1133.svg\" alt=\"$x \\in I$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.52ex; vertical-align: -0.46ex; \" src=\"img/img1225.svg\" alt=\"$x_{0}$\" loading=\"lazy\"></span> é um ponto de <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img529.svg\" alt=\"$I$\" loading=\"lazy\"></span>, então o PVI\n </i><!-- MATH\n \\begin{displaymath}\n \\left\\{ \\begin{array}{l}\n a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + \\cdots + a_{2}(x) y'' + a_{1}(x) y' + a_{0}(x) y = g(x), \\\\\n y^{(n-1)}(x_{0}) = y_{n-1}, \\quad \\dots, \\quad y''(x_{0})=y_{2}, \\quad y'(x_{0}) = y_{1}, \\quad y(x_{0}) = y_{0}\n \\end{array} \\right. \n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P87\" title=\"3P87\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1224.svg\" alt=\"$\\displaystyle \\left\\{ \\begin{array}{l}\n a_{n}(x) y^{(n)} + a_{n-1}(x) y^{(n-1)}...\n ... = y_{2}, \\quad y'(x_{0}) = y_{1}, \\quad y(x_{0}) = y_{0}\n \\end{array} \\right. $\" loading=\"lazy\">\n </div><i>\n possui uma única solução <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1201.svg\" alt=\"$y = f(x)$\" loading=\"lazy\"></span>, neste intervalo.\n </i></div>\n \n <p class=\" unidade\" id=\"3P88\" title=\"3P88\">\n Agora estamos prontos para estabelecer as identidades mencionadas no início deste capítulo. Para isto, consideremos\n primeiro o problema de valor inicial, definido em <!-- MATH\n $I = \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1236.svg\" alt=\"$I = \\mathbb{R}$\" loading=\"lazy\"></span>,\n <!-- MATH\n \\begin{displaymath}\n \\left\\{ \\begin{array}{l}\n y'' - y = 0 \\\\\n y'(0) = 0, \\quad y(0) = 1.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P89\" title=\"3P89\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1237.svg\" alt=\"$\\displaystyle \\left\\{ \\begin{array}{l}\n y'' - y = 0 \\\\\n y'(0) = 0, \\quad y(0) = 1.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P90\" title=\"3P90\">\n Notemos que a função\n <!-- MATH\n \\begin{displaymath}\n y_{1} = \\frac{1}{2} e^{x} + \\frac{1}{2} e^{-x} = \\frac{e^{x} + e^{-x}}{2}\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P91\" title=\"3P91\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1238.svg\" alt=\"$\\displaystyle y_{1} = \\frac{1}{2} e^{x} + \\frac{1}{2} e^{-x} = \\frac{e^{x} + e^{-x}}{2} $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P92\" style=\"text-indent: 0 !important;\" title=\"3P92\">\n é solução do PVI dado. Mas, do que vimos nos capítulos anteriores a respeito das funções trigonométricas hiperbólicas,\n a função <!-- MATH\n $y_{2} = \\cosh(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1239.svg\" alt=\"$y_{2} = \\cosh(x)$\" loading=\"lazy\"></span>, satisfaz a equação diferencial, pois\n <!-- MATH\n \\begin{displaymath}\n (y_{2})'' - y_{2} = (\\cosh(x))'' - \\cosh(x) = \\cosh(x) - \\cosh(x) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P93\" title=\"3P93\">\n <img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1240.svg\" alt=\"$\\displaystyle (y_{2})'' - y_{2} = (\\cosh(x))'' - \\cosh(x) = \\cosh(x) - \\cosh(x) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P94\" style=\"text-indent: 0 !important;\" title=\"3P94\">\n para todo <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> real e, além disso, <!-- MATH\n $y_{2} = \\cosh(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1239.svg\" alt=\"$y_{2} = \\cosh(x)$\" loading=\"lazy\"></span> satisfaz as duas condições iniciais,\n <!-- MATH\n \\begin{displaymath}\n y_{2}(0) = \\cosh(0) = 1, \\qquad \\text{e} \\qquad y_{2}'(0) = {\\mathrm{senh}}(0) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P95\" title=\"3P95\">\n <img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1241.svg\" alt=\"$\\displaystyle y_{2}(0) = \\cosh(0) = 1,$\" loading=\"lazy\"> e<img style=\"height: 2.45ex; vertical-align: -0.63ex; \" src=\"img/img1242.svg\" alt=\"$\\displaystyle \\qquad y_{2}'(0) = {\\mathrm{senh}}(0) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P96\" style=\"text-indent: 0 !important;\" title=\"3P96\">\n donde temos que <!-- MATH\n $y_{2} = \\cosh(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1239.svg\" alt=\"$y_{2} = \\cosh(x)$\" loading=\"lazy\"></span> é também uma solução do PVI. Mas o Teorema de Picard, garante que a solução deste\n PVI é única e, portanto, as duas soluções coincidem para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, isto é,\n <!-- MATH\n \\begin{displaymath}\n \\cosh(x) = y_{2} = y_{1} = \\frac{e^{x} + e^{-x}}{2},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P97\" title=\"3P97\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1243.svg\" alt=\"$\\displaystyle \\cosh(x) = y_{2} = y_{1} = \\frac{e^{x} + e^{-x}}{2}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P98\" style=\"text-indent: 0 !important;\" title=\"3P98\">\n qualquer que seja <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P99\" title=\"3P99\">\n Para a segunda fórmula em (<a href=\"#idexpo\">3.1</a>), consideremos outro problema de valor inicial, também definido em <!-- MATH\n $I = \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1236.svg\" alt=\"$I = \\mathbb{R}$\" loading=\"lazy\"></span>,\n <!-- MATH\n \\begin{displaymath}\n \\left\\{ \\begin{array}{l}\n y'' - y = 0 \\\\\n y'(0) = 1, \\quad y(0) = 0.\n \\end{array} \\right. \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P100\" title=\"3P100\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1244.svg\" alt=\"$\\displaystyle \\left\\{ \\begin{array}{l}\n y'' - y = 0 \\\\\n y'(0) = 1, \\quad y(0) = 0.\n \\end{array} \\right. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P101\" title=\"3P101\">\n Observe que, comparando com o PVI anterior, apenas trocamos as condições iniciais. Nestes termos, a função\n <!-- MATH\n \\begin{displaymath}\n y_{1} = \\frac{1}{2} e^{x} - \\frac{1}{2} e^{-x} = \\frac{e^{x} - e^{-x}}{2},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P102\" title=\"3P102\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1245.svg\" alt=\"$\\displaystyle y_{1} = \\frac{1}{2} e^{x} - \\frac{1}{2} e^{-x} = \\frac{e^{x} - e^{-x}}{2}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P103\" style=\"text-indent: 0 !important;\" title=\"3P103\">\n é solução deste novo PVI. Entretanto, do que vimos nos capítulos anteriores, a função <!-- MATH\n $y_{2} = {\\mathrm{senh}}(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1246.svg\" alt=\"$y_{2} = {\\mathrm{senh}}(x)$\" loading=\"lazy\"></span> também\n satisfaz a equação diferencial,\n <!-- MATH\n \\begin{displaymath}\n (y_{2})'' - y_{2} = ({\\mathrm{senh}}(x))'' - {\\mathrm{senh}}(x) = {\\mathrm{senh}}(x) - {\\mathrm{senh}}(x) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P104\" title=\"3P104\">\n <img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1247.svg\" alt=\"$\\displaystyle (y_{2})'' - y_{2} = ({\\mathrm{senh}}(x))'' - {\\mathrm{senh}}(x) = {\\mathrm{senh}}(x) - {\\mathrm{senh}}(x) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P105\" style=\"text-indent: 0 !important;\" title=\"3P105\">\n para todo <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> real e as duas condições inicias,\n <!-- MATH\n \\begin{displaymath}\n y_{2}(0) = {\\mathrm{senh}}(0) = 0, \\qquad \\text{e} \\qquad y_{2}'(0) = \\cosh(0) = 1,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P106\" title=\"3P106\">\n <img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1248.svg\" alt=\"$\\displaystyle y_{2}(0) = {\\mathrm{senh}}(0) = 0,$\" loading=\"lazy\"> e<img style=\"height: 2.45ex; vertical-align: -0.63ex; \" src=\"img/img1249.svg\" alt=\"$\\displaystyle \\qquad y_{2}'(0) = \\cosh(0) = 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P107\" style=\"text-indent: 0 !important;\" title=\"3P107\">\n e, dessa forma, <!-- MATH\n $y_{2} = {\\mathrm{senh}}(x)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1246.svg\" alt=\"$y_{2} = {\\mathrm{senh}}(x)$\" loading=\"lazy\"></span> também é solução do PVI, para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Do Teorema de Picard, segue que as\n duas soluções coincidem para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, isto é,\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{senh}}(x) = y_{2} = y_{1} = \\frac{e^{x} - e^{-x}}{2},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P108\" title=\"3P108\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1250.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}(x) = y_{2} = y_{1} = \\frac{e^{x} - e^{-x}}{2}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P109\" style=\"text-indent: 0 !important;\" title=\"3P109\">\n para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <a id=\"secforlog\"></a>\n\n:::\n\n## 3.3 As fórmulas logarítmicas {#SECTION00730000000000000000}\n \n::: {.raw_html}\n \n <p class=\" unidade\" id=\"3P110\" title=\"3P110\">\n Esta seção é dedicada à obtenção das igualdades logarítmicas em (<a href=\"#idlog\">3.2</a>), além das igualdades correspondentes às\n outras quatro funções trigonométricas hiperbólicas inversas.\n </p>\n <p class=\" unidade\" id=\"3P111\" title=\"3P111\">\n Dados <!-- MATH\n $x, y \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.04ex; vertical-align: -0.50ex; \" src=\"img/img1251.svg\" alt=\"$x, y \\in \\mathbb{R}$\" loading=\"lazy\"></span> de forma que <!-- MATH\n $y = {\\mathrm{senh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1252.svg\" alt=\"$y = {\\mathrm{senh}}^{-1} x$\" loading=\"lazy\"></span>, já sabemos que é válida a relação\n <!-- MATH\n \\begin{displaymath}\n x = {\\mathrm{senh}}y = \\frac{e^{y} - e^{-y}}{2}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P112\" title=\"3P112\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1253.svg\" alt=\"$\\displaystyle x = {\\mathrm{senh}}y = \\frac{e^{y} - e^{-y}}{2}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P113\" title=\"3P113\">\n Vamos isolar <span class=\"MATH\"><img style=\"height: 1.56ex; vertical-align: -0.50ex; \" src=\"img/img71.svg\" alt=\"$y$\" loading=\"lazy\"></span> no segundo membro e obter uma expressão para <span class=\"MATH\"><img style=\"height: 1.56ex; vertical-align: -0.50ex; \" src=\"img/img71.svg\" alt=\"$y$\" loading=\"lazy\"></span> em termos da variável independente <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span>. A igualdade\n anterior, nos leva a\n <!-- MATH\n \\begin{displaymath}\n 2x = e^{y} - e^{-y}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P114\" title=\"3P114\">\n <img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1254.svg\" alt=\"$\\displaystyle 2x = e^{y} - e^{-y}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P115\" title=\"3P115\">\n Multiplicando ambos os membros por <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> e reorganizando os termos temos\n <!-- MATH\n \\begin{displaymath}\n (e^{y})^{2} - 2xe^{y} - 1 = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P116\" title=\"3P116\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1256.svg\" alt=\"$\\displaystyle (e^{y})^{2} - 2xe^{y} - 1 = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P117\" style=\"text-indent: 0 !important;\" title=\"3P117\">\n que é uma equação quadrática na expressão <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span>. As soluções desta equação quadrática, são dadas por\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{2x \\pm \\sqrt{4x^{2} + 4} }{2} = x \\pm \\sqrt{x^{2} + 1}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P118\" title=\"3P118\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1257.svg\" alt=\"$\\displaystyle e^{y} = \\frac{2x \\pm \\sqrt{4x^{2} + 4} }{2} = x \\pm \\sqrt{x^{2} + 1}.$\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P119\" title=\"3P119\">\n Temos que descartar uma das soluções porque o lado esquerdo da igualdade acima é sempre positivo e o termo <!-- MATH\n $x -\n \\sqrt{x^{2}+1}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1258.svg\" alt=\"$x -\n \\sqrt{x^{2}+1}$\" loading=\"lazy\"></span> é sempre negativo já que <!-- MATH\n $\\sqrt{x^{2}+1} > x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1259.svg\" alt=\"$\\sqrt{x^{2}+1} > x$\" loading=\"lazy\"></span>. Tomando então a solução positiva temos <!-- MATH\n $e^{y} = x +\n \\sqrt{x^{2} + 1}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1260.svg\" alt=\"$e^{y} = x +\n \\sqrt{x^{2} + 1}$\" loading=\"lazy\"></span> e aplicando logaritmo (natural) em ambos os membros,\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{senh}}^{-1} x = y = \\ln(x + \\sqrt{x^{2} + 1}),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P120\" title=\"3P120\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1261.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}^{-1} x = y = \\ln(x + \\sqrt{x^{2} + 1}), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P121\" style=\"text-indent: 0 !important;\" title=\"3P121\">\n para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P122\" title=\"3P122\">\n Para o cosseno hiperbólico inverso, consideramos <!-- MATH\n $y = \\cosh^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1262.svg\" alt=\"$y = \\cosh^{-1} x$\" loading=\"lazy\"></span> e a relação inversa <!-- MATH\n $x = \\cosh y$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1263.svg\" alt=\"$x = \\cosh y$\" loading=\"lazy\"></span>, válida para\n todos <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1264.svg\" alt=\"$x \\geq 1$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1265.svg\" alt=\"$y \\geq 0$\" loading=\"lazy\"></span>. Como antes, tomemos a identidade\n <!-- MATH\n \\begin{displaymath}\n x = \\cosh y = \\frac{e^{y} + e^{-y}}{2},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P123\" title=\"3P123\">\n <img style=\"height: 4.72ex; vertical-align: -1.55ex; \" src=\"img/img1266.svg\" alt=\"$\\displaystyle x = \\cosh y = \\frac{e^{y} + e^{-y}}{2}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P124\" style=\"text-indent: 0 !important;\" title=\"3P124\">\n e vamos isolar <span class=\"MATH\"><img style=\"height: 1.56ex; vertical-align: -0.50ex; \" src=\"img/img71.svg\" alt=\"$y$\" loading=\"lazy\"></span> no segundo membro. De forma análoga ao caso anterior, multiplicamos os dois membros por <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1267.svg\" alt=\"$2e^{y}$\" loading=\"lazy\"></span>,\n reorganizamos os termos e chegamos a\n <!-- MATH\n \\begin{displaymath}\n (e^{y})^{2} - 2xe^{y} + 1 = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P125\" title=\"3P125\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1268.svg\" alt=\"$\\displaystyle (e^{y})^{2} - 2xe^{y} + 1 = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P126\" style=\"text-indent: 0 !important;\" title=\"3P126\">\n e resolvendo esta equação quadrática em termos de <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> temos\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{2x \\pm \\sqrt{4x^{2} - 4}}{2} = x \\pm \\sqrt{x^{2}-1}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P127\" title=\"3P127\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1269.svg\" alt=\"$\\displaystyle e^{y} = \\frac{2x \\pm \\sqrt{4x^{2} - 4}}{2} = x \\pm \\sqrt{x^{2}-1}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P128\" title=\"3P128\">\n Observamos agora que os dois termos a que se refere o segundo membro são positivos e, portanto, não há impossibilidades\n matemáticas para aplicar o logaritmo. Entretanto lembremos que <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1265.svg\" alt=\"$y \\geq 0$\" loading=\"lazy\"></span> e isto siginifica que <!-- MATH\n $e^{y} \\geq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.95ex; vertical-align: -0.38ex; \" src=\"img/img1270.svg\" alt=\"$e^{y} \\geq 1$\" loading=\"lazy\"></span>. Mas\n para <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1264.svg\" alt=\"$x \\geq 1$\" loading=\"lazy\"></span> temos que\n <!-- MATH\n \\begin{displaymath}\n x = 1+\\sqrt{(x-1)^{2}} = 1+\\sqrt{x^{2}-2x+1} \\leq 1+\\sqrt{x^{2}-2+1} = 1+\\sqrt{x^{2}-1},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P129\" title=\"3P129\">\n <img style=\"height: 4.07ex; vertical-align: -1.22ex; \" src=\"img/img1271.svg\" alt=\"$\\displaystyle x = 1+\\sqrt{(x-1)^{2}} = 1+\\sqrt{x^{2}-2x+1} \\leq 1+\\sqrt{x^{2}-2+1} = 1+\\sqrt{x^{2}-1}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P130\" style=\"text-indent: 0 !important;\" title=\"3P130\">\n e desta forma <!-- MATH\n $x - \\sqrt{x^{2}-1} \\leq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 2.46ex; vertical-align: -0.38ex; \" src=\"img/img1272.svg\" alt=\"$x - \\sqrt{x^{2}-1} \\leq 1$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P131\" title=\"3P131\">\n Descartando esta inconsistência, tomamos <!-- MATH\n $e^{y} = x + \\sqrt{x^{2}-1}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1273.svg\" alt=\"$e^{y} = x + \\sqrt{x^{2}-1}$\" loading=\"lazy\"></span> e aplicando o logaritmo em ambos os membros, temos\n <!-- MATH\n \\begin{displaymath}\n \\cosh^{-1} x = y = \\ln(x + \\sqrt{x^{2}-1}).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P132\" title=\"3P132\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1274.svg\" alt=\"$\\displaystyle \\cosh^{-1} x = y = \\ln(x + \\sqrt{x^{2}-1}). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P133\" title=\"3P133\">\n Já provamos as duas fórmulas indicadas no início deste capítulo. Contudo, vamos completar o trabalho e obter as\n fórmulas em termos do logaritmo para as demais funções trigonométricas hiperbólicas.\n </p>\n <p class=\" unidade\" id=\"3P134\" title=\"3P134\">\n Consideremos <!-- MATH\n $y = {\\mathrm {tgh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.59ex; vertical-align: -0.58ex; \" src=\"img/img1275.svg\" alt=\"$y = {\\mathrm {tgh}}^{-1} x$\" loading=\"lazy\"></span> e a relação inversa\n <!-- MATH\n \\begin{displaymath}\n x = {\\mathrm {tgh}}y = \\frac{e^{y} - e^{-y}}{e^{y} + e^{-y}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P135\" title=\"3P135\">\n <img style=\"height: 4.90ex; vertical-align: -1.72ex; \" src=\"img/img1276.svg\" alt=\"$\\displaystyle x = {\\mathrm {tgh}}y = \\frac{e^{y} - e^{-y}}{e^{y} + e^{-y}}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P136\" style=\"text-indent: 0 !important;\" title=\"3P136\">\n válida para <!-- MATH\n $x \\in (-1,1)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1277.svg\" alt=\"$x \\in (-1,1)$\" loading=\"lazy\"></span> e <!-- MATH\n $y \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.04ex; vertical-align: -0.50ex; \" src=\"img/img1278.svg\" alt=\"$y \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Organizando os termos temos\n <!-- MATH\n \\begin{displaymath}\n xe^{y} + xe^{-y} = e^{y} - e^{-y},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P137\" title=\"3P137\">\n <img style=\"height: 2.13ex; vertical-align: -0.27ex; \" src=\"img/img1279.svg\" alt=\"$\\displaystyle xe^{y} + xe^{-y} = e^{y} - e^{-y}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P138\" style=\"text-indent: 0 !important;\" title=\"3P138\">\n e multiplicando ambos os membros por <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> e reorganizando em forma de equação quadrática, chegamos a\n <!-- MATH\n \\begin{displaymath}\n (1-x)(e^{y})^{2} - (1+x) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P139\" title=\"3P139\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1280.svg\" alt=\"$\\displaystyle (1-x)(e^{y})^{2} - (1+x) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P140\" style=\"text-indent: 0 !important;\" title=\"3P140\">\n que resolvida em termos de <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> fornece\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\pm \\sqrt{ \\tfrac{1+x}{1-x} } = \\pm \\left( \\tfrac{1+x}{1-x} \\right)^{\\frac{1}{2}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P141\" title=\"3P141\">\n <img style=\"height: 4.08ex; vertical-align: -1.28ex; \" src=\"img/img1281.svg\" alt=\"$\\displaystyle e^{y} = \\pm \\sqrt{ \\tfrac{1+x}{1-x} } = \\pm \\left( \\tfrac{1+x}{1-x} \\right)^{\\frac{1}{2}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P142\" title=\"3P142\">\n Como o primeiro membro é sempre positivo, então descartamos a solução negativa. Observemos também que como <!-- MATH\n $x \\in\n (-1,1)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1277.svg\" alt=\"$x \\in (-1,1)$\" loading=\"lazy\"></span> então a fração dentro da raiz quadrada é sempre positiva, o que não acarreta mais inconsistências. Aplicando\n então o logaritmo, temos que\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {tgh}}^{-1} x = y = \\ln \\left( \\tfrac{1+x}{1-x} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{1+x}{1-x} \\right),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P143\" title=\"3P143\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1282.svg\" alt=\"$\\displaystyle {\\mathrm {tgh}}^{-1} x = y = \\ln \\left( \\tfrac{1+x}{1-x} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{1+x}{1-x} \\right), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P144\" style=\"text-indent: 0 !important;\" title=\"3P144\">\n para todo <!-- MATH\n $x \\in (-1,1)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1277.svg\" alt=\"$x \\in (-1,1)$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P145\" title=\"3P145\">\n Agora a cotangente hiperbólica inversa. Tomamos <!-- MATH\n $y = {\\mathrm{ctgh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.59ex; vertical-align: -0.58ex; \" src=\"img/img1283.svg\" alt=\"$y = {\\mathrm{ctgh}}^{-1} x$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $x \\in (-\\infty,-1) \\cup (1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1284.svg\" alt=\"$x \\in (-\\infty,-1) \\cup (1,\\infty)$\" loading=\"lazy\"></span>, com\n <!-- MATH\n $y \\in \\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1285.svg\" alt=\"$y \\in \\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span> e então\n <!-- MATH\n \\begin{displaymath}\n x = {\\mathrm{ctgh}}y = \\frac{e^{y} + e^{-y}}{e^{y} - e^{-y}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P146\" title=\"3P146\">\n <img style=\"height: 4.90ex; vertical-align: -1.72ex; \" src=\"img/img1286.svg\" alt=\"$\\displaystyle x = {\\mathrm{ctgh}}y = \\frac{e^{y} + e^{-y}}{e^{y} - e^{-y}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P147\" title=\"3P147\">\n Como no caso da tangente hiperbólica, reorganizamos os termos e multiplicamos por <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> ambos os membros e chegamos a\n <!-- MATH\n \\begin{displaymath}\n (x-1)(e^{y})^{2} - (x+1) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P148\" title=\"3P148\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1287.svg\" alt=\"$\\displaystyle (x-1)(e^{y})^{2} - (x+1) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P149\" style=\"text-indent: 0 !important;\" title=\"3P149\">\n e resolvendo esta equação quadrática em <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> temos\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\pm \\sqrt{ \\tfrac{x+1}{x-1} } = \\pm \\left( \\tfrac{x+1}{x-1} \\right)^{\\frac{1}{2}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P150\" title=\"3P150\">\n <img style=\"height: 4.08ex; vertical-align: -1.28ex; \" src=\"img/img1288.svg\" alt=\"$\\displaystyle e^{y} = \\pm \\sqrt{ \\tfrac{x+1}{x-1} } = \\pm \\left( \\tfrac{x+1}{x-1} \\right)^{\\frac{1}{2}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P151\" title=\"3P151\">\n Note que a fração dentro da raiz quadrada é sempre positiva para <!-- MATH\n $x \\in (-\\infty,-1) \\cup (1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1284.svg\" alt=\"$x \\in (-\\infty,-1) \\cup (1,\\infty)$\" loading=\"lazy\"></span>. De fato, o\n numerador e o denominador são ambos negativos no intervalo <!-- MATH\n $(-\\infty,-1)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1289.svg\" alt=\"$(-\\infty,-1)$\" loading=\"lazy\"></span> e são ambos positivos no intervalo\n <!-- MATH\n $(1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1290.svg\" alt=\"$(1,\\infty)$\" loading=\"lazy\"></span>. Vamos descartar a solução negativa, pois o lado esquerdo da igualdade é sempre positivo. Assim, tomando a\n solução positiva e aplicando logaritmo em ambos os membros, vem\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{ctgh}}^{-1} x = y = \\ln \\left( \\tfrac{x+1}{x-1} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{x+1}{x-1} \\right).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P152\" title=\"3P152\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1291.svg\" alt=\"$\\displaystyle {\\mathrm{ctgh}}^{-1} x = y = \\ln \\left( \\tfrac{x+1}{x-1} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{x+1}{x-1} \\right). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P153\" title=\"3P153\">\n Para a secante hiperbólica inversa, fazendo <!-- MATH\n $y = {\\mathrm{sech}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1292.svg\" alt=\"$y = {\\mathrm{sech}}^{-1} x$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $x \\in (0,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1293.svg\" alt=\"$x \\in (0,1]$\" loading=\"lazy\"></span>, com <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1265.svg\" alt=\"$y \\geq 0$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n x = {\\mathrm{sech}}y = \\frac{2}{e^{y} + e^{-y}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P154\" title=\"3P154\">\n <img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1294.svg\" alt=\"$\\displaystyle x = {\\mathrm{sech}}y = \\frac{2}{e^{y} + e^{-y}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P155\" title=\"3P155\">\n Após reorganização dos termos e multiplicação por <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span>, obtemos a equação quadrática\n <!-- MATH\n \\begin{displaymath}\n x(e^{y})^{2} - 2e^{y} + x = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P156\" title=\"3P156\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1295.svg\" alt=\"$\\displaystyle x(e^{y})^{2} - 2e^{y} + x = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P157\" style=\"text-indent: 0 !important;\" title=\"3P157\">\n que resolvida em termos de <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span> nos traz\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{2 \\pm \\sqrt{4 - 4x^{2}} }{2x} = \\frac{1 \\pm \\sqrt{1-x^{2}} }{x}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P158\" title=\"3P158\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1296.svg\" alt=\"$\\displaystyle e^{y} = \\frac{2 \\pm \\sqrt{4 - 4x^{2}} }{2x} = \\frac{1 \\pm \\sqrt{1-x^{2}} }{x}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P159\" title=\"3P159\">\n Notemos que para <!-- MATH\n $x \\in (0,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1293.svg\" alt=\"$x \\in (0,1]$\" loading=\"lazy\"></span> ocorre <!-- MATH\n $1-x^{2} \\geq 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.31ex; vertical-align: -0.38ex; \" src=\"img/img1297.svg\" alt=\"$1-x^{2} \\geq 0$\" loading=\"lazy\"></span> e, portanto, não temos problemas com a raiz quadrada. Entretanto,\n para <!-- MATH\n $x \\in (0,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1293.svg\" alt=\"$x \\in (0,1]$\" loading=\"lazy\"></span> temos <!-- MATH\n $(1-x) \\geq 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1298.svg\" alt=\"$(1-x) \\geq 0$\" loading=\"lazy\"></span> e então\n <!-- MATH\n \\begin{displaymath}\n 1-x = \\sqrt{(1-x)^{2}} = \\sqrt{(1-x)(1-x)} \\leq \\sqrt{(1-x)(1+x)} = \\sqrt{1-x^{2}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P160\" title=\"3P160\">\n <img style=\"height: 4.07ex; vertical-align: -1.22ex; \" src=\"img/img1299.svg\" alt=\"$\\displaystyle 1-x = \\sqrt{(1-x)^{2}} = \\sqrt{(1-x)(1-x)} \\leq \\sqrt{(1-x)(1+x)} = \\sqrt{1-x^{2}}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P161\" style=\"text-indent: 0 !important;\" title=\"3P161\">\n donde temos que <!-- MATH\n $1 - \\sqrt{1-x^{2}} \\leq x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.46ex; vertical-align: -0.38ex; \" src=\"img/img1300.svg\" alt=\"$1 - \\sqrt{1-x^{2}} \\leq x$\" loading=\"lazy\"></span> e, portanto, <!-- MATH\n $\\frac{1 - \\sqrt{1-x^{2}} }{x} \\leq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 3.22ex; vertical-align: -0.83ex; \" src=\"img/img1301.svg\" alt=\"$\\frac{1 - \\sqrt{1-x^{2}} }{x} \\leq 1$\" loading=\"lazy\"></span>. Mas isto é inconsistente\n com o primeiro membro <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span>, que é maior ou igual a 1, já que <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1265.svg\" alt=\"$y \\geq 0$\" loading=\"lazy\"></span>. Só não seria inconsistente caso os dois\n termos fossem iguais a 1, isto é <!-- MATH\n $e^{y} = 1 = \\frac{1 - \\sqrt{1-x^{2}} }{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.22ex; vertical-align: -0.83ex; \" src=\"img/img1302.svg\" alt=\"$e^{y} = 1 = \\frac{1 - \\sqrt{1-x^{2}} }{x}$\" loading=\"lazy\"></span>, que somente ocorre se <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1303.svg\" alt=\"$y = 0$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.11ex; \" src=\"img/img1304.svg\" alt=\"$x = 1$\" loading=\"lazy\"></span>.\n Mas a igualdade <!-- MATH\n $e^{y} = \\frac{1 + \\sqrt{1-x^{2}} }{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.22ex; vertical-align: -0.83ex; \" src=\"img/img1305.svg\" alt=\"$e^{y} = \\frac{1 + \\sqrt{1-x^{2}} }{x}$\" loading=\"lazy\"></span> também se verifica para <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1303.svg\" alt=\"$y = 0$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.11ex; \" src=\"img/img1304.svg\" alt=\"$x = 1$\" loading=\"lazy\"></span> e, portanto, podemos\n descartar totalmente a solução <!-- MATH\n $e^{y} = \\frac{1-\\sqrt{1-x^{2}} }{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.22ex; vertical-align: -0.83ex; \" src=\"img/img1306.svg\" alt=\"$e^{y} = \\frac{1-\\sqrt{1-x^{2}} }{x}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P162\" title=\"3P162\">\n Tomando então a solução que não apresenta inconsistências, tomamos\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1 + \\sqrt{1-x^{2}} }{x},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P163\" title=\"3P163\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1307.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1 + \\sqrt{1-x^{2}} }{x}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P164\" style=\"text-indent: 0 !important;\" title=\"3P164\">\n e aplicando o logaritmo, temos\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{sech}}^{-1} x = y = \\ln \\left( \\frac{1}{x} + \\frac{\\sqrt{1-x^{2}}}{x} \\right).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P165\" title=\"3P165\">\n <img style=\"height: 6.51ex; vertical-align: -2.74ex; \" src=\"img/img1308.svg\" alt=\"$\\displaystyle {\\mathrm{sech}}^{-1} x = y = \\ln \\left( \\frac{1}{x} + \\frac{\\sqrt{1-x^{2}}}{x} \\right). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P166\" title=\"3P166\">\n Finalmente, considerando <!-- MATH\n $y = {\\mathrm{csch}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.51ex; vertical-align: -0.50ex; \" src=\"img/img1309.svg\" alt=\"$y = {\\mathrm{csch}}^{-1} x$\" loading=\"lazy\"></span> e a relação inversa\n <!-- MATH\n \\begin{displaymath}\n x = {\\mathrm{csch}}y = \\frac{2}{e^{y} - e^{-y}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P167\" title=\"3P167\">\n <img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1310.svg\" alt=\"$\\displaystyle x = {\\mathrm{csch}}y = \\frac{2}{e^{y} - e^{-y}}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P168\" style=\"text-indent: 0 !important;\" title=\"3P168\">\n válida para todos <!-- MATH\n $x \\in \\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1311.svg\" alt=\"$x \\in \\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span> e <!-- MATH\n $y \\in \\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1285.svg\" alt=\"$y \\in \\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span>. Procedendo como no caso da secante, obtemos a equação quadrática\n <!-- MATH\n \\begin{displaymath}\n x(e^{y})^{2} - 2e^{y} - x = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P169\" title=\"3P169\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1312.svg\" alt=\"$\\displaystyle x(e^{y})^{2} - 2e^{y} - x = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P170\" style=\"text-indent: 0 !important;\" title=\"3P170\">\n que resolvida em <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1255.svg\" alt=\"$e^{y}$\" loading=\"lazy\"></span>, nos fornece\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{2 \\pm \\sqrt{4 + 4x^{2}} }{2x} = \\frac{1 \\pm \\sqrt{1 + x^{2}} }{x}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P171\" title=\"3P171\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1313.svg\" alt=\"$\\displaystyle e^{y} = \\frac{2 \\pm \\sqrt{4 + 4x^{2}} }{2x} = \\frac{1 \\pm \\sqrt{1 + x^{2}} }{x}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P172\" title=\"3P172\">\n Notemos que como antes, queremos que o membro da direita seja positivo, pois o da esquerda o é. O termo\n <!-- MATH\n $\\sqrt{1+x^{2}}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1314.svg\" alt=\"$\\sqrt{1+x^{2}}$\" loading=\"lazy\"></span> é sempre maior que 1. O numerador assume portanto valores positivos considerando <!-- MATH\n $1 + \\sqrt{1+x^{2}}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1315.svg\" alt=\"$1 + \\sqrt{1+x^{2}}$\" loading=\"lazy\"></span>,\n e valores negativos considerando <!-- MATH\n $1 - \\sqrt{1+x^{2}}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.38ex; vertical-align: -0.31ex; \" src=\"img/img1316.svg\" alt=\"$1 - \\sqrt{1+x^{2}}$\" loading=\"lazy\"></span>. Mas como <!-- MATH\n $x \\in \\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1311.svg\" alt=\"$x \\in \\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span> temos que o denominador também assume\n valores positivos e valores negativos. Então se <span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img1317.svg\" alt=\"$x < 0$\" loading=\"lazy\"></span>, devemos considerar a solução\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1 - \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} - \\frac{\\sqrt{1+x^{2}}}{x} > 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P173\" title=\"3P173\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1318.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1 - \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} - \\frac{\\sqrt{1+x^{2}}}{x} > 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P174\" style=\"text-indent: 0 !important;\" title=\"3P174\">\n e se <span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img637.svg\" alt=\"$x>0$\" loading=\"lazy\"></span>, devemos considerar a solução\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1 + \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} + \\frac{\\sqrt{1+x^{2}}}{x} > 0.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P175\" title=\"3P175\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1319.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1 + \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} + \\frac{\\sqrt{1+x^{2}}}{x} > 0. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P176\" title=\"3P176\">\n Podemos ainda obter uma única expressão válida para os dois casos. Observe que se <span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img637.svg\" alt=\"$x>0$\" loading=\"lazy\"></span> podemos escrever\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1}{x} + \\frac{ \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} + \\frac{ \\sqrt{1 + x^{2}} }{|x|},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P177\" title=\"3P177\">\n <img style=\"height: 5.57ex; vertical-align: -2.07ex; \" src=\"img/img1320.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1}{x} + \\frac{ \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} + \\frac{ \\sqrt{1 + x^{2}} }{\\vert x\\vert}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P178\" style=\"text-indent: 0 !important;\" title=\"3P178\">\n e se <span class=\"MATH\"><img style=\"height: 1.69ex; vertical-align: -0.14ex; \" src=\"img/img1317.svg\" alt=\"$x < 0$\" loading=\"lazy\"></span>, podemos escrever\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1}{x} - \\frac{ \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} + \\frac{\\sqrt{1 + x^{2}}}{-x} = \\frac{1}{x} + \\frac{\\sqrt{1 + x^{2}}}{|x|}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P179\" title=\"3P179\">\n <img style=\"height: 5.57ex; vertical-align: -2.07ex; \" src=\"img/img1321.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1}{x} - \\frac{ \\sqrt{1 + x^{2}} }{x} = \\frac{1}{x} ...\n ...c{\\sqrt{1 + x^{2}}}{-x} = \\frac{1}{x} + \\frac{\\sqrt{1 + x^{2}}}{\\vert x\\vert}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P180\" title=\"3P180\">\n Assim, para qualquer <!-- MATH\n $x \\in \\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1311.svg\" alt=\"$x \\in \\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span>, escrevemos\n <!-- MATH\n \\begin{displaymath}\n e^{y} = \\frac{1}{x} + \\frac{\\sqrt{1 + x^{2}} }{|x|},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P181\" title=\"3P181\">\n <img style=\"height: 5.57ex; vertical-align: -2.07ex; \" src=\"img/img1322.svg\" alt=\"$\\displaystyle e^{y} = \\frac{1}{x} + \\frac{\\sqrt{1 + x^{2}} }{\\vert x\\vert}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P182\" style=\"text-indent: 0 !important;\" title=\"3P182\">\n e aplicando o logaritmo, temos\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{csch}}^{-1} x = y = \\ln \\left( \\tfrac{1}{x} + \\tfrac{\\sqrt{1 + x^{2}} }{|x|} \\right),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P183\" title=\"3P183\">\n <img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1323.svg\" alt=\"$\\displaystyle {\\mathrm{csch}}^{-1} x = y = \\ln \\left( \\tfrac{1}{x} + \\tfrac{\\sqrt{1 + x^{2}} }{\\vert x\\vert} \\right), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P184\" style=\"text-indent: 0 !important;\" title=\"3P184\">\n que é válida para todo <!-- MATH\n $x \\in \\mathbb{R}^{*}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.75ex; vertical-align: -0.18ex; \" src=\"img/img1324.svg\" alt=\"$x \\in \\mathbb{R}^{*}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P185\" title=\"3P185\">\n A tabela abaixo reúne as fórmulas desta seção.\n \n </p>\n <div class=\"CENTER\"><a id=\"3995\"></a>\n <table>\n <caption><strong>Tabela 3.1:</strong>\n Fórmulas logarítmicas para as funções trigonométricas hiperbólicas inversas.</caption>\n <tbody><tr><td>\n <div class=\"CENTER\">\n <table class=\"PAD BORDER\">\n <tbody><tr><td class=\"LEFT\">função</td>\n <td class=\"CENTER\">domínio</td>\n <td class=\"CENTER\">igualdade logarítmica</td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n ${\\mathrm{senh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.14ex; vertical-align: -0.13ex; \" src=\"img/img1325.svg\" alt=\"${\\mathrm{senh}}^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.64ex; vertical-align: -0.10ex; \" src=\"img/img217.svg\" alt=\"$\\mathbb{R}$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\ln(x + \\sqrt{x^{2} + 1})$\n -->\n <span class=\"MATH\"><img style=\"height: 2.70ex; vertical-align: -0.62ex; \" src=\"img/img1326.svg\" alt=\"$\\ln(x + \\sqrt{x^{2} + 1})$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n $\\cosh^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.14ex; vertical-align: -0.13ex; \" src=\"img/img1327.svg\" alt=\"$\\cosh^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $[1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img547.svg\" alt=\"$[1,\\infty)$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\ln(x + \\sqrt{x^{2}-1})$\n -->\n <span class=\"MATH\"><img style=\"height: 2.70ex; vertical-align: -0.62ex; \" src=\"img/img1328.svg\" alt=\"$\\ln(x + \\sqrt{x^{2}-1})$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n ${\\mathrm {tgh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.59ex; vertical-align: -0.58ex; \" src=\"img/img1329.svg\" alt=\"${\\mathrm {tgh}}^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img614.svg\" alt=\"$(-1,1)$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\frac{1}{2} \\ln \\left( \\tfrac{1+x}{1-x} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.94ex; vertical-align: -0.95ex; \" src=\"img/img1330.svg\" alt=\"$\\frac{1}{2} \\ln \\left( \\tfrac{1+x}{1-x} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n ${\\mathrm{ctgh}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.59ex; vertical-align: -0.58ex; \" src=\"img/img1331.svg\" alt=\"${\\mathrm{ctgh}}^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <!-- MATH\n $\\mathbb{R}- [-1,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1332.svg\" alt=\"$\\mathbb{R}- [-1,1]$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $\\frac{1}{2} \\ln \\left( \\tfrac{x+1}{x-1} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.94ex; vertical-align: -0.95ex; \" src=\"img/img1333.svg\" alt=\"$\\frac{1}{2} \\ln \\left( \\tfrac{x+1}{x-1} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n ${\\mathrm{sech}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.14ex; vertical-align: -0.13ex; \" src=\"img/img1334.svg\" alt=\"${\\mathrm{sech}}^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img887.svg\" alt=\"$(0,1]$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\ln \\left( \\frac{1}{x} + \\frac{\\sqrt{1-x^{2}}}{x} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1335.svg\" alt=\"$\\ln \\left( \\frac{1}{x} + \\frac{\\sqrt{1-x^{2}}}{x} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"LEFT\"><!-- MATH\n ${\\mathrm{csch}}^{-1} x$\n -->\n <span class=\"MATH\"><img style=\"height: 2.14ex; vertical-align: -0.13ex; \" src=\"img/img1336.svg\" alt=\"${\\mathrm{csch}}^{-1} x$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\mathbb{R}-\\{0\\}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1337.svg\" alt=\"$\\mathbb{R}-\\{0\\}$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"><!-- MATH\n $\\ln \\left( \\tfrac{1}{x} + \\tfrac{\\sqrt{1 + x^{2}} }{|x|} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1338.svg\" alt=\"$\\ln \\left( \\tfrac{1}{x} + \\tfrac{\\sqrt{1 + x^{2}} }{\\vert x\\vert} \\right)$\" loading=\"lazy\"></span></td>\n </tr>\n </tbody></table>\n <a id=\"tabdfhi\"></a>\n </div></td></tr>\n </tbody></table>\n </div>\n <br>\n \n <p class=\" unidade\" id=\"3P186\" title=\"3P186\">\n Note que as fórmulas de derivação das funções trigonométricas hiperbólicas inversas, foram obtidas na seção\n <a href=\"/trigonometria-hiperbolica/funcoes-trigonometricas-hiperbolicas#secderhipinv\">2.8</a> e resumidas na tabela <a href=\"/trigonometria-hiperbolica/funcoes-trigonometricas-hiperbolicas#tabderhipinv\">2.4</a>. Naquela seção foi utilizado o método da diferenciação\n implícita. As fórmulas de derivação da tabela <a href=\"/trigonometria-hiperbolica/funcoes-trigonometricas-hiperbolicas#tabderhipinv\">2.4</a> podem também ser obtidas derivando diretamente as\n expressões logarítmicas da tabela <a href=\"#tabdfhi\">3.1</a>. Deixamos os detalhes para o leitor.\n </p>\n\n:::\n\n## 3.4 Extensão às variáveis complexas {#SECTION00740000000000000000}\n \n::: {.raw_html} \n <p class=\" unidade\" id=\"3P187\" title=\"3P187\">\n Identidades similares das identidades (<a href=\"#idexpo\">3.1</a>) são conhecidas para as funções trigonométricas circulares. Mas\n isto exigirá o uso de números complexos. Além disso, modelos matemáticos que representam fenômenos físicos são\n constantemente usados para estudar e conhecer esses fenômenos — e em várias situações — a representação desses fenômenos\n exige a utilização de números complexos juntamente com funções trigonométricas. Em virtude disso, apresentaremos nesta\n seção como são definidas as funções trigonométricas circulares e hiperbólicas de uma variável complexa.\n </p>\n <p class=\" unidade\" id=\"3P188\" title=\"3P188\">\n Usando as séries de potências das funções trigonométricas, desenvolvidas na seção anterior, vamos construir as funções\n trigonométricas de variáveis complexas. Na seção <a href=\"#secspot\">3.1</a>, vimos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P189\" title=\"3P189\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1339.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1340.svg\" alt=\"$\\displaystyle = x - \\frac{1}{3!} x^{3} + \\frac{1}{5!} x^{5} - \\frac{1}{7!} x^{7} + \\frac{1}{9!} x^{9} - \\frac{1}{11!} x^{11} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1341.svg\" alt=\"$\\displaystyle \\cos x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1342.svg\" alt=\"$\\displaystyle = 1 - \\frac{1}{2!} x^{2} + \\frac{1}{4!} x^{4} - \\frac{1}{6!} x^{6} + \\frac{1}{8!} x^{8} - \\frac{1}{10!} x^{10} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1343.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1344.svg\" alt=\"$\\displaystyle = x + \\frac{1}{3!} x^{3} + \\frac{1}{5!} x^{5} + \\frac{1}{7!} x^{7} + \\frac{1}{9!} x^{9} + \\frac{1}{11!} x^{11} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1345.svg\" alt=\"$\\displaystyle \\cosh x$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1346.svg\" alt=\"$\\displaystyle = 1 + \\frac{1}{2!} x^{2} + \\frac{1}{4!} x^{4} + \\frac{1}{6!} x^{6} + \\frac{1}{8!} x^{8} + \\frac{1}{10!} x^{10} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P190\" style=\"text-indent: 0 !important;\" title=\"3P190\">\n para todo <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P191\" title=\"3P191\">\n Observe que o lado direito destas igualdades faz sentido se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> for um número complexo, desde que a série seja\n convergente para este número complexo. Isto nos sugere que a igualdade possa ser utilizada para definir as funções\n trigonométricas seno e cosseno para os números complexos que tornam a série convergente. Nestes termos, se <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span>,\n então definimos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P192\" title=\"3P192\"><table>\n <tbody><tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1348.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1349.svg\" alt=\"$\\displaystyle = z - \\frac{1}{3!} z^{3} + \\frac{1}{5!} z^{5} - \\frac{1}{7!} z^{7} + \\frac{1}{9!} z^{9} - \\frac{1}{11!} z^{11} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"senz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">12</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1350.svg\" alt=\"$\\displaystyle \\cos z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1351.svg\" alt=\"$\\displaystyle = 1 - \\frac{1}{2!} z^{2} + \\frac{1}{4!} z^{4} - \\frac{1}{6!} z^{6} + \\frac{1}{8!} z^{8} - \\frac{1}{10!} z^{10} +\n \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"cosz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">13</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1352.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1353.svg\" alt=\"$\\displaystyle = z + \\frac{1}{3!} z^{3} + \\frac{1}{5!} z^{5} + \\frac{1}{7!} z^{7} + \\frac{1}{9!} z^{9} + \\frac{1}{11!} z^{11} + \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"senhz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">14</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1354.svg\" alt=\"$\\displaystyle \\cosh z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1355.svg\" alt=\"$\\displaystyle = 1 - \\frac{1}{2!} z^{2} + \\frac{1}{4!} z^{4} + \\frac{1}{6!} z^{6} + \\frac{1}{8!} z^{8} + \\frac{1}{10!} z^{10} +\n \\cdots,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"coshz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">15</span>)</a></td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P193\" style=\"text-indent: 0 !important;\" title=\"3P193\">\n desde que as séries convirjam.\n </p>\n <p class=\" unidade\" id=\"3P194\" title=\"3P194\">\n Precisamos determinar os valores <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span> que tornam estas séries convergentes. Para isto, recorremos ao teste da\n razão (Critério de D'Alembert), para garantir a convergência de séries de potências de variável complexa. A\n demonstração deste teorema pode ser encontrada em algum texto de Variáveis complexas. Recomendamos [<a href=\"/trigonometria-hiperbolica/referencias#ZillVC\">11</a>, Zill].\n \n </p>\n <div id=\"3Teo9\" title=\"3Teo9\" class=\"unidade\"><a id=\"DAlembert\"><b>Teorema <span class=\"arabic\">3</span>.<span class=\"arabic\">9</span></b></a> (Teste da razão) \n <i>Se <!-- MATH\n $\\{z_{n}\\}_{n \\in \\mathbb{N}}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1356.svg\" alt=\"$\\{z_{n}\\}_{n \\in \\mathbb{N}}$\" loading=\"lazy\"></span> é uma sequência de números complexos e\n </i><!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left| \\frac{z_{n+1}}{z_{n}} \\right| = L < 1,\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P195\" title=\"3P195\">\n <img style=\"height: 5.24ex; vertical-align: -2.10ex; \" src=\"img/img1357.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert \\frac{z_{n+1}}{z_{n}} \\right\\vert = L < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P196\" style=\"text-indent: 0 !important;\" title=\"3P196\"><i>\n então a série <!-- MATH\n $\\sum z_{n}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1358.svg\" alt=\"$\\sum z_{n}$\" loading=\"lazy\"></span> é absolutamente convergente (e, portanto, convergente).\n </i></p></div>\n \n <p class=\" unidade\" id=\"3P197\" title=\"3P197\">\n A série de potências (<a href=\"#senz\">3.12</a>) converge qualquer que seja <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span>, pois\n <!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left| - \\frac{\\frac{1}{(2n+1)!} z^{2n+1}}{\\frac{1}{(2n-1)!} z^{2n-1}} \\right|\n =\\lim_{n \\to \\infty} \\left| \\frac{(2n-1)!}{(2n+1)!} \\frac{z^{2n+1}}{z^{2n-1}} \\right| = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} |z^{2}| = 0 < 1, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P198\" title=\"3P198\">\n <img style=\"height: 6.84ex; vertical-align: -2.91ex; \" src=\"img/img1359.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert - \\frac{\\frac{1}{(2n+1)!} z^{2n+1}...\n ...\\right\\vert = \\lim_{n \\to \\infty} \\frac{1}{(2n+1)2n} \\vert z^{2}\\vert = 0 < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P199\" style=\"text-indent: 0 !important;\" title=\"3P199\">\n para qualquer <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span>. Analogamente para as séries de potências em (<a href=\"#cosz\">3.13</a>), (<a href=\"#senhz\">3.14</a>) e (<a href=\"#coshz\">3.15</a>).\n </p>\n <p class=\" unidade\" id=\"3P200\" title=\"3P200\">\n Naturalmente as definições (<a href=\"#senz\">3.12</a>)-(<a href=\"#coshz\">3.15</a>) não são muito cômodas para trabalharmos. Vamos então tentar\n modificar estas expressões para redefinir seno e cosseno de números complexos em termos de funções reais de variável\n real. Independente de modificarmos estas expressões, os membros na direita destas igualdades são números complexos e\n então o que esperamos é que possamos reescrever a série de potências como sendo um número complexo mais simples de ser\n manipulado, dado na forma tradicional <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1360.svg\" alt=\"$a + bi$\" loading=\"lazy\"></span> com <!-- MATH\n $a, b \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1361.svg\" alt=\"$a, b \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P201\" title=\"3P201\">\n Comecemos então com a identidade (<a href=\"#senz\">3.12</a>), colocando <!-- MATH\n $z = x + yi$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1362.svg\" alt=\"$z = x + yi$\" loading=\"lazy\"></span>, com <!-- MATH\n $x, y \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.04ex; vertical-align: -0.50ex; \" src=\"img/img1251.svg\" alt=\"$x, y \\in \\mathbb{R}$\" loading=\"lazy\"></span>. Temos então\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P202\" title=\"3P202\"><a id=\"eqseno\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm {sen}}(x+yi) = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (x+yi)^{2n+1}.\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1363.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}(x+yi) = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (x+yi)^{2n+1}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">16</span>)\n </td></tr>\n\n </tbody></table>\n\n \n \n </div>\n \n \n <p class=\" unidade\" id=\"3P203\" title=\"3P203\">\n Usando a fórmula da expansão binomial para o termo <!-- MATH\n $(x+yi)^{2n+1}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.55ex; vertical-align: -0.62ex; \" src=\"img/img1364.svg\" alt=\"$(x+yi)^{2n+1}$\" loading=\"lazy\"></span>, podemos reescrever (<a href=\"#eqseno\">3.16</a>) como\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P204\" title=\"3P204\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1365.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}(x+yi)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1366.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (x+yi)^{2n+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1367.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} \\sum_{r=0}^{2n+1} \\frac{(2n+1)!}{r!(2n-r+1)!} x^{2n-r+1} (yi)^{r}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1368.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{r!(2n-r+1)!} x^{2n-r+1} (yi)^{r}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P205\" title=\"3P205\">\n O próximo lema será útil para trabalhar com o somatório duplo do segundo membro desta última igualdade.\n \n </p>\n <div id=\"3Teo10\" title=\"3Teo10\" class=\"unidade\"><b>Lema <span class=\"arabic\">3</span>.<span class=\"arabic\">10</span></b> \n <i>Para qualquer <!-- MATH\n $m \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1369.svg\" alt=\"$m \\in \\mathbb{N}$\" loading=\"lazy\"></span>,\n </i>\n <div class=\"mathdisplay unidade\" id=\"3P206\" title=\"3P206\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1370.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.16ex; vertical-align: -2.07ex; \" src=\"img/img1371.svg\" alt=\"$\\displaystyle \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1372.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1373.svg\" alt=\"$\\displaystyle \\qquad \\qquad - \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m+2)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m+2}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n </div>\n \n \n <div><i>Prova</i>.\n \n Tomando\n <!-- MATH\n \\begin{displaymath}\n \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m},\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P207\" title=\"3P207\">\n <img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1374.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P208\" style=\"text-indent: 0 !important;\" title=\"3P208\">\n vamos separar o caso <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1375.svg\" alt=\"$n = 0$\" loading=\"lazy\"></span> do somatório externo e depois os casos <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1376.svg\" alt=\"$r = 0$\" loading=\"lazy\"></span> do somatório interno. Desta forma, para\n qualquer que seja <!-- MATH\n $m \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1369.svg\" alt=\"$m \\in \\mathbb{N}$\" loading=\"lazy\"></span>, obtemos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P209\" title=\"3P209\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1377.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1378.svg\" alt=\"$\\displaystyle \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 0.19ex; vertical-align: -0.10ex; \" src=\"img/img1379.svg\" alt=\"$\\displaystyle %\n $\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1380.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} x (yi)^{m} + \\frac{1}{(m+1)!} (yi)^{m+1} + \\sum_{n...\n ...infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1381.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} x (yi)^{m} + \\frac{1}{(m+1)!} (yi)^{m+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1382.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\left( \\frac{(-1)^{n}}{m!(2n+1)!} x^...\n ...sum_{r=1}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m} \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.49ex; vertical-align: -2.90ex; \" src=\"img/img1383.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} x (yi)^{m} + \\frac{1}{(m+1)!} (yi)^{m+1} + \\sum_{n=1}^{\\infty} \\frac{(-1)^{n}}{m!(2n+1)!} x^{2n+1} (yi)^{m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1384.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\sum_{r=1}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.67ex; vertical-align: -2.90ex; \" src=\"img/img1385.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} \\left( x + \\sum_{n=1}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} x^{2n+1} \\right) + \\frac{1}{(m+1)!} (yi)^{m+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1384.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\sum_{r=1}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1386.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi...\n ...nfty} \\sum_{r=1}^{2n+1}\n \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P210\" title=\"3P210\">\n Separando novamente os temos em <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.11ex; \" src=\"img/img1387.svg\" alt=\"$r = 1$\" loading=\"lazy\"></span> do somatório interno, temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P211\" title=\"3P211\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1377.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1378.svg\" alt=\"$\\displaystyle \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 0.19ex; vertical-align: -0.10ex; \" src=\"img/img1379.svg\" alt=\"$\\displaystyle %\n $\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1388.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi...\n ...infty} \\sum_{r=1}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1389.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1390.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\left( \\frac{(-1)^{n}}{(m+1)!(2n)!} ...\n ...\\sum_{r=2}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}\\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.49ex; vertical-align: -2.90ex; \" src=\"img/img1391.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1} + \\sum_{n=1}^{\\infty} \\frac{(-1)^{n}}{(m+1)!(2n)!} x^{2n} (yi)^{m+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1392.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\sum_{r=2}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.67ex; vertical-align: -2.90ex; \" src=\"img/img1393.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1} \\left( 1 + \\sum_{n=1}^{\\infty} \\frac{(-1)^{n}}{(2n)!} x^{2n} \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.89ex; vertical-align: -2.90ex; \" src=\"img/img1392.svg\" alt=\"$\\displaystyle \\qquad + \\sum_{n=1}^{\\infty} \\sum_{r=2}^{2n+1} \\frac{(-1)^{n}}{(r+m)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1372.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1394.svg\" alt=\"$\\displaystyle \\qquad - \\sum_{n=0}^{\\infty} \\sum_{r=2}^{2n+3} \\frac{(-1)^{n}}{(r+m)!(2n-r+3)!} x^{2n-r+3} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1372.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} {\\mathrm {sen}}(x) + \\frac{1}{(m+1)!} (yi)^{m+1} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1395.svg\" alt=\"$\\displaystyle \\qquad - \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+m+2)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+m+2},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P212\" style=\"text-indent: 0 !important;\" title=\"3P212\">\n exatamente como desejado.\n <span style=\"float: right\"><img style=\"height: 1.59ex; vertical-align: -0.10ex; \" src=\"img/img193.svg\" alt=\"$\\qedsymbol$\" loading=\"lazy\"></span>\n </p></div>\n \n <p class=\" unidade\" id=\"3P213\" title=\"3P213\">\n Usando agora repetidamente este lema temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P214\" title=\"3P214\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right; vertical-align: middle;\"><span class=\"MATH\">sen</span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1397.svg\" alt=\"$\\displaystyle (x+yi) = \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{r!(2n-r+1)!} x^{2n-r+1} (yi)^{r}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1398.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) + (yi)\\cos(x) - \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+2)!(2n-r+1)!} x^{2n-r+1}(yi)^{r+2}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1399.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) + (yi)\\cos(x) - \\frac{1}{2!} (yi)^{2} {\\mathrm {sen}}(x) - \\frac{1}{3!} (yi)^{3} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1400.svg\" alt=\"$\\displaystyle \\qquad \\qquad + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+4)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+4}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1401.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) + (yi)\\cos(x) - \\frac{1}{2!} (yi)^{2} {\\math...\n ...s(x) + \\frac{1}{4!} (yi)^{4} {\\mathrm {sen}}(x) + \\frac{1}{5!} (yi)^{5} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1402.svg\" alt=\"$\\displaystyle \\qquad \\qquad + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+6)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+6}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1401.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) + (yi)\\cos(x) - \\frac{1}{2!} (yi)^{2} {\\math...\n ...s(x) + \\frac{1}{4!} (yi)^{4} {\\mathrm {sen}}(x) + \\frac{1}{5!} (yi)^{5} \\cos(x)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1403.svg\" alt=\"$\\displaystyle \\qquad \\qquad - \\frac{1}{6!} (yi)^{6} {\\mathrm {sen}}(x) - \\frac{...\n ...} \\sum_{r=0}^{2n+1} \\frac{(-1)^{n}}{(r+8)!(2n-r+1)!} x^{2n-r+1} (yi)^{r+8}, %\n $\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P215\" style=\"text-indent: 0 !important;\" title=\"3P215\">\n e assim sucessivamente. Desta forma, obtemos\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {sen}}(x+yi) = {\\mathrm {sen}}(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n)!}(yi)^{2n}\n + \\cos(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (yi)^{2n+1}, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P216\" title=\"3P216\">\n <img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1404.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}(x+yi) = {\\mathrm {sen}}(x) \\sum_{n=0}^{\\infty} \\f...\n ...(yi)^{2n}\n + \\cos(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (yi)^{2n+1}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P217\" style=\"text-indent: 0 !important;\" title=\"3P217\">\n e usando o fato de que <!-- MATH\n $(yi)^{2n} = y^{2n}i^{2n} = (-1)^{n} y^{2n}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.55ex; vertical-align: -0.62ex; \" src=\"img/img1405.svg\" alt=\"$(yi)^{2n} = y^{2n}i^{2n} = (-1)^{n} y^{2n}$\" loading=\"lazy\"></span> e que <!-- MATH\n $(yi)^{2n+1} = y^{2n+1}i^{2n+1} = (-1)^{n} y^{2n+1} i$\n -->\n <span class=\"MATH\"><img style=\"height: 2.55ex; vertical-align: -0.62ex; \" src=\"img/img1406.svg\" alt=\"$(yi)^{2n+1} = y^{2n+1}i^{2n+1} = (-1)^{n} y^{2n+1} i$\" loading=\"lazy\"></span>, então temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P218\" title=\"3P218\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1365.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}(x+yi)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1407.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n)!}(yi)^{2n} + \\cos(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (yi)^{2n+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1408.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n)!}(-...\n ...y^{2n} + \\cos(x) \\sum_{n=0}^{\\infty} \\frac{(-1)^{n}}{(2n+1)!} (-1)^{n}y^{2n+1}i$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1409.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) \\sum_{n=0}^{\\infty} \\frac{1}{(2n)!}y^{2n} + i \\cos(x) \\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)!}y^{2n+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1410.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(x) \\cosh(y) + i \\cos(x) {\\mathrm{senh}}(y).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P219\" title=\"3P219\">\n Temos portanto uma definição alternativa e mais elegante para a definição do seno de um número complexo <!-- MATH\n $z = x + yi$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1362.svg\" alt=\"$z = x + yi$\" loading=\"lazy\"></span>.\n Definição esta que será também útil para os nossos propósitos. Não estamos interessados em repetir o procedimento\n anterior, mas ele pode ser aplicado também às funções cosseno, senho hiperbólico e cosseno hiperbólico para obter\n expressões mais simples. Como não repetiremos o processo anterior apenas enunciaremos as expressões finais na próxima definição.\n </p>\n <div><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">11</span></b> \n Dado <!-- MATH\n $z = u + iv \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1411.svg\" alt=\"$z = u + iv \\in \\mathbb{C}$\" loading=\"lazy\"></span>, os números complexos seno de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span>, cosseno de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span>, seno hiperbólico de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span> e cosseno\n hiperbólico de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span>, são dados respectivamente por,\n \n <div class=\"mathdisplay unidade\" id=\"3P220\" title=\"3P220\"><table>\n <tbody><tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1348.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1413.svg\" alt=\"$\\displaystyle = {\\mathrm {sen}}(u+iv) = {\\mathrm {sen}}u \\cosh v + i{\\mathrm{senh}}v \\cos u,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"defsinz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">17</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1350.svg\" alt=\"$\\displaystyle \\cos z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1414.svg\" alt=\"$\\displaystyle = \\cos(u+iv) = \\cos u \\cosh v - i{\\mathrm {sen}}u {\\mathrm{senh}}v,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"defcosz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">18</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1352.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1415.svg\" alt=\"$\\displaystyle = {\\mathrm{senh}}(u+iv) = {\\mathrm{senh}}u \\cos v + i{\\mathrm {sen}}v \\cosh u,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right; vertical-align: middle;\">\n <a id=\"defsenhz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">19</span>)</a></td></tr>\n <tr>\n <td style=\"text-align:center; vertical-align: middle;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1354.svg\" alt=\"$\\displaystyle \\cosh z$\" loading=\"lazy\"></span>\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1416.svg\" alt=\"$\\displaystyle = \\cosh(u+iv) = \\cosh u \\cos v + i{\\mathrm{senh}}u {\\mathrm {sen}}v.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n <a id=\"defcoshz\">(<span class=\"arabic\">3</span>.<span class=\"arabic\">20</span>)</a></td></tr>\n </tbody></table></div>\n </div>\n \n <p class=\" unidade\" id=\"3P221\" title=\"3P221\">\n Vamos analisar um pouco mais estas funções e verificar que elas possuem propriedades similares às funções\n trigonométricas com argumentos reais. É natural esperar por isto, pois extensões não devem desorganizar o que já estava\n “funcionando”. Comecemos com os casos circulares.\n </p>\n <p class=\" unidade\" id=\"3P222\" title=\"3P222\">\n Vamos determinar as raízes das funções seno e cosseno. Queremos então determinar os valores de <!-- MATH\n $z = u + iv \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1411.svg\" alt=\"$z = u + iv \\in \\mathbb{C}$\" loading=\"lazy\"></span> para\n os quais <!-- MATH\n ${\\mathrm {sen}}z = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1417.svg\" alt=\"${\\mathrm {sen}}z = 0$\" loading=\"lazy\"></span>. Nestes termos queremos determinar os valores (reais) de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> tais que\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {sen}}z = {\\mathrm {sen}}u \\cosh v + i {\\mathrm{senh}}v \\cos u = 0.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P223\" title=\"3P223\">\n <img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1418.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}z = {\\mathrm {sen}}u \\cosh v + i {\\mathrm{senh}}v \\cos u = 0. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P224\" title=\"3P224\">\n Da igualdade de números complexos temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P225\" title=\"3P225\"><table class=\"equation\">\n <tbody><tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1419.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}u \\cosh v = 0,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1420.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}v \\cos u = 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P226\" title=\"3P226\">\n Da primeira equação, como <!-- MATH\n $\\cosh v \\geq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1421.svg\" alt=\"$\\cosh v \\geq 1$\" loading=\"lazy\"></span> para todo <!-- MATH\n $v \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1422.svg\" alt=\"$v \\in \\mathbb{R}$\" loading=\"lazy\"></span>, então resta que <!-- MATH\n ${\\mathrm {sen}}u = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img117.svg\" alt=\"${\\mathrm {sen}}u = 0$\" loading=\"lazy\"></span>. Temos assim que <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img216.svg\" alt=\"$u = k \\pi$\" loading=\"lazy\"></span> para qualquer <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Com estes valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> na segunda equação temos que <!-- MATH\n $\\cos u = \\pm 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1423.svg\" alt=\"$\\cos u = \\pm 1$\" loading=\"lazy\"></span> e então resta\n que <!-- MATH\n ${\\mathrm{senh}}v = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1424.svg\" alt=\"${\\mathrm{senh}}v = 0$\" loading=\"lazy\"></span> donde obtemos <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1425.svg\" alt=\"$v = 0$\" loading=\"lazy\"></span>. Assim,\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {sen}}z = 0 \\qquad \\text{se e somente se} \\qquad z = u+iv = k\\pi,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P227\" title=\"3P227\">\n <img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1426.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}z = 0$\" loading=\"lazy\"> se e somente se<img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1427.svg\" alt=\"$\\displaystyle \\qquad z = u+iv = k\\pi, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P228\" style=\"text-indent: 0 !important;\" title=\"3P228\">\n para qualquer <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>, exatamente como no caso real.\n </p>\n <p class=\" unidade\" id=\"3P229\" title=\"3P229\">\n Analogamente para determinar os valores de <!-- MATH\n $z = u + iv \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1411.svg\" alt=\"$z = u + iv \\in \\mathbb{C}$\" loading=\"lazy\"></span> tais que <!-- MATH\n $\\cos z = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1428.svg\" alt=\"$\\cos z = 0$\" loading=\"lazy\"></span>, temos que encontrar os valores\n reais de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> tais que\n <!-- MATH\n \\begin{displaymath}\n \\cos z = \\cos u \\cosh v - i {\\mathrm {sen}}u {\\mathrm{senh}}v = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P230\" title=\"3P230\">\n <img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1429.svg\" alt=\"$\\displaystyle \\cos z = \\cos u \\cosh v - i {\\mathrm {sen}}u {\\mathrm{senh}}v = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P231\" style=\"text-indent: 0 !important;\" title=\"3P231\">\n e da igualdade de números complexos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P232\" title=\"3P232\"><table class=\"equation\">\n <tbody><tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1430.svg\" alt=\"$\\displaystyle \\cos u \\cosh v = 0,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1431.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}u {\\mathrm{senh}}v = 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P233\" title=\"3P233\">\n Como <!-- MATH\n $\\cosh v \\geq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1421.svg\" alt=\"$\\cosh v \\geq 1$\" loading=\"lazy\"></span> para todo <!-- MATH\n $v \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1422.svg\" alt=\"$v \\in \\mathbb{R}$\" loading=\"lazy\"></span>, da primeira equação resta que <!-- MATH\n $\\cos u = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img131.svg\" alt=\"$\\cos u = 0$\" loading=\"lazy\"></span> e então <!-- MATH\n $u = \\frac{\\pi}{2} + k\n \\pi$\n -->\n <span class=\"MATH\"><img style=\"height: 2.45ex; vertical-align: -0.83ex; \" src=\"img/img222.svg\" alt=\"$u = \\frac{\\pi}{2} + k\\pi$\" loading=\"lazy\"></span> para qualquer <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Como <!-- MATH\n ${\\mathrm {sen}}u = \\pm 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1432.svg\" alt=\"${\\mathrm {sen}}u = \\pm 1$\" loading=\"lazy\"></span>, substituindo na segunda equação vem <!-- MATH\n ${\\mathrm{senh}}v = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1424.svg\" alt=\"${\\mathrm{senh}}v = 0$\" loading=\"lazy\"></span> e, portanto, <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1425.svg\" alt=\"$v = 0$\" loading=\"lazy\"></span>. Assim,\n <!-- MATH\n \\begin{displaymath}\n \\cos z = 0 \\qquad \\text{se e somente se} \\qquad z = u+iv = \\frac{\\pi}{2} + k\\pi,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P234\" title=\"3P234\">\n <img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1433.svg\" alt=\"$\\displaystyle \\cos z = 0$\" loading=\"lazy\"> se e somente se<img style=\"height: 4.03ex; vertical-align: -1.55ex; \" src=\"img/img1434.svg\" alt=\"$\\displaystyle \\qquad z = u+iv = \\frac{\\pi}{2} + k\\pi, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P235\" style=\"text-indent: 0 !important;\" title=\"3P235\">\n para qualquer <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>, também como no caso real.\n </p>\n <p class=\" unidade\" id=\"3P236\" title=\"3P236\">\n Dessa forma, as demais funções trigonométricas circulares com argumentos complexos são definidas, em termos destas duas,\n como no caso de variável real, respeitando o domínio de definição. São portanto,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P237\" title=\"3P237\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.58ex; \" src=\"img/img1435.svg\" alt=\"$\\displaystyle {\\mathrm {tg}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.06ex; vertical-align: -1.57ex; \" src=\"img/img1436.svg\" alt=\"$\\displaystyle = \\frac{{\\mathrm {sen}}z}{\\cos z}$\" loading=\"lazy\"> para todo<img style=\"height: 4.03ex; vertical-align: -1.55ex; \" src=\"img/img1437.svg\" alt=\"$\\displaystyle \\quad z \\neq \\frac{\\pi}{2} + k \\pi \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.58ex; \" src=\"img/img1438.svg\" alt=\"$\\displaystyle {\\mathrm {ctg}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.06ex; vertical-align: -1.57ex; \" src=\"img/img1439.svg\" alt=\"$\\displaystyle = \\frac{\\cos z}{{\\mathrm {sen}}z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1440.svg\" alt=\"$\\displaystyle \\quad z \\neq k \\pi \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1441.svg\" alt=\"$\\displaystyle \\sec z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1442.svg\" alt=\"$\\displaystyle = \\frac{1}{\\cos z}$\" loading=\"lazy\"> para todo<img style=\"height: 4.03ex; vertical-align: -1.55ex; \" src=\"img/img1437.svg\" alt=\"$\\displaystyle \\quad z \\neq \\frac{\\pi}{2} + k \\pi \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1443.svg\" alt=\"$\\displaystyle \\csc z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1444.svg\" alt=\"$\\displaystyle = \\frac{1}{{\\mathrm {sen}}z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1445.svg\" alt=\"$\\displaystyle \\quad z \\neq k \\pi \\quad (k \\in \\mathbb{Z}).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P238\" title=\"3P238\">\n Podemos facilmente verificar, pelas igualdades (<a href=\"#defsinz\">3.17</a>) e (<a href=\"#defcosz\">3.18</a>), a validade para o caso complexo de\n identidades conhecidas para o caso real, tais como\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P239\" title=\"3P239\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1446.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}^{2}z + \\cos^{2}z = 1,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1447.svg\" alt=\"$\\displaystyle \\cos(z+\\pi) = -\\cos z$\" loading=\"lazy\"> e<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1448.svg\" alt=\"$\\displaystyle \\qquad {\\mathrm {sen}}(z+\\pi) = -{\\mathrm {sen}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1449.svg\" alt=\"$\\displaystyle \\cos(-z) = \\cos z$\" loading=\"lazy\"> e<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1450.svg\" alt=\"$\\displaystyle \\qquad {\\mathrm {sen}}(-z) = -{\\mathrm {sen}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.61ex; vertical-align: -0.58ex; \" src=\"img/img1451.svg\" alt=\"$\\displaystyle 1+{\\mathrm {tg}}^{2}z = \\sec^{2}z$\" loading=\"lazy\"> e<img style=\"height: 2.61ex; vertical-align: -0.58ex; \" src=\"img/img1452.svg\" alt=\"$\\displaystyle \\qquad 1+{\\mathrm {ctg}}^{2}z = \\csc^{2}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P240\" style=\"text-indent: 0 !important;\" title=\"3P240\">\n dentre muitas outras.\n </p>\n <p class=\" unidade\" id=\"3P241\" title=\"3P241\">\n Uma consequência direta da definição das funções seno e cosseno por série de potência é que estas funções são\n analíticas no domínio de convergência da série, isto é, no plano complexo todo. Sendo assim, estas funções satisfazem\n as condições de Cauchy-Riemann em todo o plano complexo e isto nos dá uma forma rápida para determinar as derivadas\n destas duas funções. Para um estudo mais aprofundado sobre funções analíticas e as condições de Cauchy-Riemann pode-se\n consultar algum texto sobre variáveis complexas. Recomendamos [<a href=\"/trigonometria-hiperbolica/referencias#ZillVC\">11</a>, Zill].\n </p>\n <p class=\" unidade\" id=\"3P242\" title=\"3P242\">\n Por hora basta saber que se <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1453.svg\" alt=\"$z = u+iv$\" loading=\"lazy\"></span> e <!-- MATH\n $f(z) = g(u,v)+ih(u,v)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1454.svg\" alt=\"$f(z) = g(u,v)+ih(u,v)$\" loading=\"lazy\"></span> é uma função analítica em uma região do plano\n complexo, então\n <!-- MATH\n \\begin{displaymath}\n f'(z) = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u} = \\frac{\\partial h}{\\partial v} - i \\frac{\\partial g}{\\partial v},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P243\" title=\"3P243\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1455.svg\" alt=\"$\\displaystyle f'(z) = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u} = \\frac{\\partial h}{\\partial v} - i \\frac{\\partial g}{\\partial v}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P244\" style=\"text-indent: 0 !important;\" title=\"3P244\">\n para todo <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span> nesta região.\n </p>\n <p class=\" unidade\" id=\"3P245\" title=\"3P245\">\n Considerando <!-- MATH\n $f(z) = {\\mathrm {sen}}z = g(u,v) + i h(u,v)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1456.svg\" alt=\"$f(z) = {\\mathrm {sen}}z = g(u,v) + i h(u,v)$\" loading=\"lazy\"></span>, temos <!-- MATH\n $g(u,v) = {\\mathrm {sen}}u \\cosh v$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1457.svg\" alt=\"$g(u,v) = {\\mathrm {sen}}u \\cosh v$\" loading=\"lazy\"></span> e <!-- MATH\n $h(u,v) = {\\mathrm{senh}}v \\cos u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1458.svg\" alt=\"$h(u,v) = {\\mathrm{senh}}v \\cos u$\" loading=\"lazy\"></span> e então\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P246\" title=\"3P246\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1459.svg\" alt=\"$\\displaystyle f'(z) = ({\\mathrm {sen}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1460.svg\" alt=\"$\\displaystyle = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1461.svg\" alt=\"$\\displaystyle = \\frac{\\partial}{\\partial u} ({\\mathrm {sen}}u \\cosh v) + i \\frac{\\partial}{\\partial u}({\\mathrm{senh}}v \\cos u)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1462.svg\" alt=\"$\\displaystyle = \\cos u \\cosh v - i {\\mathrm{senh}}v {\\mathrm {sen}}u = \\cos z.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P247\" title=\"3P247\">\n Também, se <!-- MATH\n $f(z) = \\cos z = g(u,v) + i h(u,v)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1463.svg\" alt=\"$f(z) = \\cos z = g(u,v) + i h(u,v)$\" loading=\"lazy\"></span> então <!-- MATH\n $g(u,v) = \\cos u \\cosh v$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1464.svg\" alt=\"$g(u,v) = \\cos u \\cosh v$\" loading=\"lazy\"></span> e <!-- MATH\n $h(u,v) = -{\\mathrm {sen}}u {\\mathrm{senh}}v$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1465.svg\" alt=\"$h(u,v) = -{\\mathrm {sen}}u {\\mathrm{senh}}v$\" loading=\"lazy\"></span> e, dessa\n forma,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P248\" title=\"3P248\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1466.svg\" alt=\"$\\displaystyle f'(z) = (\\cos z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1460.svg\" alt=\"$\\displaystyle = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1467.svg\" alt=\"$\\displaystyle = \\frac{\\partial}{\\partial u} (\\cos u \\cosh v) + i \\frac{\\partial}{\\partial u}(- {\\mathrm {sen}}u {\\mathrm{senh}}v)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1468.svg\" alt=\"$\\displaystyle = -{\\mathrm {sen}}u \\cosh v - i {\\mathrm{senh}}v \\cos u = - {\\mathrm {sen}}z.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P249\" title=\"3P249\">\n As derivadas das funções seno e cosseno de variável complexa são então respectivamente o cosseno e o oposto do seno,\n exatamente como no caso real. Já que as regras de derivação para funções complexas são as mesmas para funções reais,\n isto é,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P250\" title=\"3P250\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1469.svg\" alt=\"$\\displaystyle (f(z)+g(z))'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1470.svg\" alt=\"$\\displaystyle = f'(z) + g'(z),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1471.svg\" alt=\"$\\displaystyle (f(z) \\cdot g(z))'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1472.svg\" alt=\"$\\displaystyle = f'(z) \\cdot g(z) + f(z) \\cdot g'(z),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 5.49ex; vertical-align: -2.10ex; \" src=\"img/img1473.svg\" alt=\"$\\displaystyle \\left( \\frac{f(z)}{g(z)} \\right)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.20ex; vertical-align: -2.07ex; \" src=\"img/img1474.svg\" alt=\"$\\displaystyle = \\frac{f'(z) \\cdot g(z) - f(z) \\cdot g'(z)}{(g(z))^{2}},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P251\" style=\"text-indent: 0 !important;\" title=\"3P251\">\n então as derivadas das demais funções trigonométricas circulares, são também iguais às derivadas obtidas no caso real.\n São portanto\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P252\" title=\"3P252\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1475.svg\" alt=\"$\\displaystyle ({\\mathrm {tg}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.16ex; vertical-align: -0.13ex; \" src=\"img/img1476.svg\" alt=\"$\\displaystyle = \\sec^{2}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1477.svg\" alt=\"$\\displaystyle ({\\mathrm {ctg}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1478.svg\" alt=\"$\\displaystyle = -\\csc^{2}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1479.svg\" alt=\"$\\displaystyle (\\sec z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.58ex; \" src=\"img/img1480.svg\" alt=\"$\\displaystyle = \\sec z {\\mathrm {tg}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1481.svg\" alt=\"$\\displaystyle (\\csc z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.90ex; vertical-align: -0.58ex; \" src=\"img/img1482.svg\" alt=\"$\\displaystyle = -\\csc z {\\mathrm {ctg}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P253\" style=\"text-indent: 0 !important;\" title=\"3P253\">\n respeitados os domínios de definição.\n </p>\n <p class=\" unidade\" id=\"3P254\" title=\"3P254\">\n Além disso, as identidades obtidas <!-- MATH\n $\\cos(iu) = \\cosh u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1483.svg\" alt=\"$\\cos(iu) = \\cosh u$\" loading=\"lazy\"></span> e <!-- MATH\n ${\\mathrm {sen}}(iu) = i {\\mathrm{senh}}u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1484.svg\" alt=\"${\\mathrm {sen}}(iu) = i {\\mathrm{senh}}u$\" loading=\"lazy\"></span> permitem estabelecer uma\n correspondência entre as funções trigonométricas circulares e as suas respectivas hiperbólicas. As correspondências das\n demais funções trigonométricas ficam\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P255\" title=\"3P255\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1485.svg\" alt=\"$\\displaystyle {\\mathrm {tg}}(ui)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.16ex; vertical-align: -2.07ex; \" src=\"img/img1486.svg\" alt=\"$\\displaystyle = \\frac{{\\mathrm {sen}}(ui)}{\\cos(ui)} = \\frac{i{\\mathrm{senh}}u}{\\cosh u} = i{\\mathrm {tgh}}u$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1487.svg\" alt=\"$\\displaystyle {\\mathrm {ctg}}(ui)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.16ex; vertical-align: -2.07ex; \" src=\"img/img1488.svg\" alt=\"$\\displaystyle = \\frac{\\cos(ui)}{{\\mathrm {sen}}(ui)} = \\frac{\\cosh u}{i{\\mathrm{senh}}u} = \\frac{-i\\cosh u}{{\\mathrm{senh}}u} = -i{\\mathrm{ctgh}}u$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1489.svg\" alt=\"$\\displaystyle \\sec(ui)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1490.svg\" alt=\"$\\displaystyle = \\frac{1}{\\cos(ui)} = \\frac{1}{\\cosh u} = {\\mathrm{sech}}u$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1491.svg\" alt=\"$\\displaystyle \\csc(ui)$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1492.svg\" alt=\"$\\displaystyle = \\frac{1}{{\\mathrm {sen}}(ui)} = \\frac{1}{i{\\mathrm{senh}}u} = \\frac{-i}{{\\mathrm{senh}}u} = -i {\\mathrm{csch}}u.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P256\" title=\"3P256\">\n Vamos estudar outras propriedades das funções trigonométricas hiperbólicas de variável complexa. Comecemos pelas raízes\n destas funções. Queremos determinar os números complexos <!-- MATH\n $z = u+iv \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1411.svg\" alt=\"$z = u + iv \\in \\mathbb{C}$\" loading=\"lazy\"></span> tais que <!-- MATH\n ${\\mathrm{senh}}z = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1493.svg\" alt=\"${\\mathrm{senh}}z = 0$\" loading=\"lazy\"></span>. Nestes termos\n devemos encontrar números reais <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> tais que\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{senh}}z = {\\mathrm{senh}}u \\cos v + i{\\mathrm {sen}}v \\cosh u = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P257\" title=\"3P257\">\n <img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1494.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}z = {\\mathrm{senh}}u \\cos v + i{\\mathrm {sen}}v \\cosh u = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P258\" style=\"text-indent: 0 !important;\" title=\"3P258\">\n e da igualdade de complexos, <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> devem satisfazer\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P259\" title=\"3P259\"><table class=\"equation\">\n <tbody><tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1495.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}u \\cos v = 0,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1496.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}v \\cosh u = 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P260\" title=\"3P260\">\n Da segunda equação, como <!-- MATH\n $\\cosh u \\geq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1497.svg\" alt=\"$\\cosh u \\geq 1$\" loading=\"lazy\"></span> para todo <!-- MATH\n $u \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img57.svg\" alt=\"$u \\in \\mathbb{R}$\" loading=\"lazy\"></span>, então devemos ter <!-- MATH\n ${\\mathrm {sen}}v = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1498.svg\" alt=\"${\\mathrm {sen}}v = 0$\" loading=\"lazy\"></span> e, portanto, <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1499.svg\" alt=\"$v = k\\pi$\" loading=\"lazy\"></span>\n para <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Com estes valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> na primeira equação, resta que <!-- MATH\n ${\\mathrm{senh}}u = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1500.svg\" alt=\"${\\mathrm{senh}}u = 0$\" loading=\"lazy\"></span> e então <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img239.svg\" alt=\"$u = 0$\" loading=\"lazy\"></span>. Temos assim que\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm{senh}}z = 0 \\qquad \\text{se e somente se} \\qquad z = ik\\pi,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P261\" title=\"3P261\">\n <img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1501.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}z = 0$\" loading=\"lazy\"> se e somente se<img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1502.svg\" alt=\"$\\displaystyle \\qquad z = ik\\pi, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P262\" style=\"text-indent: 0 !important;\" title=\"3P262\">\n para <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Observe que estas raízes são complexas e que a única destas raízes que é real, é <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1503.svg\" alt=\"$z = 0$\" loading=\"lazy\"></span>, que coincide\n com a única raiz real da função seno hiperbólico a argumento real.\n </p>\n <p class=\" unidade\" id=\"3P263\" title=\"3P263\">\n Agora vamos determinar <!-- MATH\n $z = u+iv \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1411.svg\" alt=\"$z = u + iv \\in \\mathbb{C}$\" loading=\"lazy\"></span> tal que <!-- MATH\n $\\cosh z = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1504.svg\" alt=\"$\\cosh z = 0$\" loading=\"lazy\"></span>. Da identidade (<a href=\"#defcoshz\">3.20</a>), queremos determinar os\n valores reais de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> que satisfazem\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P264\" title=\"3P264\"><table class=\"equation\">\n <tbody><tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1505.svg\" alt=\"$\\displaystyle \\cosh u \\cos v = 0,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1506.svg\" alt=\"$\\displaystyle {\\mathrm{senh}}u {\\mathrm {sen}}v = 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P265\" title=\"3P265\">\n Como <!-- MATH\n $\\cosh u \\geq 1$\n -->\n <span class=\"MATH\"><img style=\"height: 1.94ex; vertical-align: -0.38ex; \" src=\"img/img1497.svg\" alt=\"$\\cosh u \\geq 1$\" loading=\"lazy\"></span> então da primeira equação segue que <!-- MATH\n $\\cos v = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1507.svg\" alt=\"$\\cos v = 0$\" loading=\"lazy\"></span> e, portanto, <!-- MATH\n $v = \\frac{\\pi}{2} + k\\pi$\n -->\n <span class=\"MATH\"><img style=\"height: 2.45ex; vertical-align: -0.83ex; \" src=\"img/img1508.svg\" alt=\"$v = \\frac{\\pi}{2} + k\\pi$\" loading=\"lazy\"></span> para <!-- MATH\n $k\n \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Com estes valores de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img2.svg\" alt=\"$v$\" loading=\"lazy\"></span> na segunda equação temos que <!-- MATH\n ${\\mathrm{senh}}u = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1500.svg\" alt=\"${\\mathrm{senh}}u = 0$\" loading=\"lazy\"></span> e então <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img239.svg\" alt=\"$u = 0$\" loading=\"lazy\"></span>. Segue que\n <!-- MATH\n \\begin{displaymath}\n \\cosh z = 0 \\qquad \\text{se e somente se} \\qquad z = i(\\tfrac{\\pi}{2}+k\\pi),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P266\" title=\"3P266\">\n <img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1509.svg\" alt=\"$\\displaystyle \\cosh z = 0$\" loading=\"lazy\"> se e somente se<img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img1510.svg\" alt=\"$\\displaystyle \\qquad z = i(\\tfrac{\\pi}{2}+k\\pi), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P267\" style=\"text-indent: 0 !important;\" title=\"3P267\">\n para <!-- MATH\n $k \\in \\mathbb{Z}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img48.svg\" alt=\"$k \\in \\mathbb{Z}$\" loading=\"lazy\"></span>. Dentre estas raízes complexas não existe nenhuma raiz real, o que ratifica a não existência de números\n reais <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> tais que <!-- MATH\n $\\cosh x = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1511.svg\" alt=\"$\\cosh x = 0$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P268\" title=\"3P268\">\n As demais funções trigonométricas hiperbólicas de variáveis complexas são definidas em termos de seno e cosseno como no\n caso real, restritas ao domínio de definição. Isto é,\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P269\" title=\"3P269\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1512.svg\" alt=\"$\\displaystyle {\\mathrm {tgh}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1513.svg\" alt=\"$\\displaystyle = \\frac{{\\mathrm{senh}}z}{\\cosh z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img1514.svg\" alt=\"$\\displaystyle \\quad z \\neq i(\\tfrac{\\pi}{2}+k\\pi) \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1515.svg\" alt=\"$\\displaystyle {\\mathrm{ctgh}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1516.svg\" alt=\"$\\displaystyle = \\frac{\\cosh z}{{\\mathrm{senh}}z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1517.svg\" alt=\"$\\displaystyle \\quad z \\neq ik\\pi \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1518.svg\" alt=\"$\\displaystyle {\\mathrm{sech}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1519.svg\" alt=\"$\\displaystyle = \\frac{1}{\\cosh z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img1514.svg\" alt=\"$\\displaystyle \\quad z \\neq i(\\tfrac{\\pi}{2}+k\\pi) \\quad (k \\in \\mathbb{Z}),$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1520.svg\" alt=\"$\\displaystyle {\\mathrm{csch}}z$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.55ex; vertical-align: -1.57ex; \" src=\"img/img1521.svg\" alt=\"$\\displaystyle = \\frac{1}{{\\mathrm{senh}}z}$\" loading=\"lazy\"> para todo<img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1522.svg\" alt=\"$\\displaystyle \\quad z \\neq ik\\pi \\quad (k \\in \\mathbb{Z}).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P270\" title=\"3P270\">\n As definições de seno e cosseno hiperbólicos em termos de séries de potências, convergentes em todo o plano complexo,\n nos diz que estas funções são analíticas em todo o plano complexo e então podemos determinar facilmente as derivadas\n destas funções. Considerando que <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1453.svg\" alt=\"$z = u+iv$\" loading=\"lazy\"></span> e que <!-- MATH\n $f(z) = {\\mathrm{senh}}z = g(u,v) + i h(u,v)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1523.svg\" alt=\"$f(z) = {\\mathrm{senh}}z = g(u,v) + i h(u,v)$\" loading=\"lazy\"></span> temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P271\" title=\"3P271\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1524.svg\" alt=\"$\\displaystyle f'(z) = ({\\mathrm{senh}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1460.svg\" alt=\"$\\displaystyle = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1525.svg\" alt=\"$\\displaystyle = \\frac{\\partial}{\\partial u}({\\mathrm{senh}}u \\cos v) + i\\frac{\\partial}{\\partial u}({\\mathrm {sen}}v \\cosh u)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1526.svg\" alt=\"$\\displaystyle = \\cosh u \\cos v + i{\\mathrm {sen}}v {\\mathrm{senh}}u = \\cosh z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P272\" style=\"text-indent: 0 !important;\" title=\"3P272\">\n exatamente como no caso de variáveis reais.\n </p>\n <p class=\" unidade\" id=\"3P273\" title=\"3P273\">\n Analogamente, para a função <!-- MATH\n $f(z) = \\cosh z = g(u,v) + ih(u,v)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1527.svg\" alt=\"$f(z) = \\cosh z = g(u,v) + ih(u,v)$\" loading=\"lazy\"></span>, temos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P274\" title=\"3P274\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1528.svg\" alt=\"$\\displaystyle f'(z) = (\\cosh z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1460.svg\" alt=\"$\\displaystyle = \\frac{\\partial g}{\\partial u} + i \\frac{\\partial h}{\\partial u}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1529.svg\" alt=\"$\\displaystyle = \\frac{\\partial}{\\partial u}(\\cosh u \\cos v) + i\\frac{\\partial}{\\partial u}({\\mathrm{senh}}u {\\mathrm {sen}}v)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1530.svg\" alt=\"$\\displaystyle = {\\mathrm{senh}}u \\cos v + i\\cosh u {\\mathrm {sen}}v = {\\mathrm{senh}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P275\" style=\"text-indent: 0 !important;\" title=\"3P275\">\n também como no caso real. Considerando ainda que a regra de derivação para o quociente de funções de variáveis\n complexas é idêntica à regra de derivação para o quociente de funções de variáveis reais, então temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P276\" title=\"3P276\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1531.svg\" alt=\"$\\displaystyle ({\\mathrm {tgh}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.16ex; vertical-align: -0.13ex; \" src=\"img/img1532.svg\" alt=\"$\\displaystyle = {\\mathrm{sech}}^{2} z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1533.svg\" alt=\"$\\displaystyle ({\\mathrm{ctgh}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1534.svg\" alt=\"$\\displaystyle = -{\\mathrm{csch}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1535.svg\" alt=\"$\\displaystyle ({\\mathrm{sech}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1536.svg\" alt=\"$\\displaystyle = -{\\mathrm{sech}}z {\\mathrm {tgh}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.44ex; vertical-align: -0.62ex; \" src=\"img/img1537.svg\" alt=\"$\\displaystyle ({\\mathrm{csch}}z)'$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.58ex; \" src=\"img/img1538.svg\" alt=\"$\\displaystyle = -{\\mathrm{csch}}z {\\mathrm{ctgh}}z,$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P277\" style=\"text-indent: 0 !important;\" title=\"3P277\">\n com a devida restrição do domínio de definição. São as mesmas fórmulas de derivação que as funções trigonométricas\n hiperbólicas de variáveis reais.\n </p>\n <p class=\" unidade\" id=\"3P278\" title=\"3P278\">\n Podem ainda ser definidas as funções trigonométricas circulares e hiperbólicas inversas a argumentos complexos. Não\n vamos nos estender neste aspecto, em virtude de que o caso complexo não é o foco do nosso interesse. Além disso,\n entraríamos no campo das funções multivalentes, isto é, funções <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1539.svg\" alt=\"$f(z)$\" loading=\"lazy\"></span> que assumem mais de um valor para cada <!-- MATH\n $z \\in\n \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span>. Esta categoria de funções foge do conceito de função de um curso de Cálculo Diferencial e Integral.\n </p>\n\n::: \n\n## 3.5 Fórmulas exponenciais para funções trigonométricas circulares {#SECTION00750000000000000000}\n\n::: {.raw_html}\n \n <p class=\" unidade\" id=\"3P279\" title=\"3P279\">\n Nesta seção, obteremos fórmulas exponenciais similares às identidades em (<a href=\"#idexpo\">3.1</a>) para as funções trigonométricas\n circulares. Mais precisamente, provaremos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P280\" title=\"3P280\"><a id=\"idexpocirc\"></a><!-- MATH\n \\begin{equation}\n {\\mathrm {sen}}x = \\frac{e^{ix}-e^{-ix}}{2i} \\qquad \\text{e} \\qquad \\cos x = \\frac{e^{ix} + e^{-ix}}{2}.\n \\end{equation}\n -->\n <table>\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 4.89ex; vertical-align: -1.55ex; \" src=\"img/img1540.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}x = \\frac{e^{ix}-e^{-ix}}{2i}$\" loading=\"lazy\"> e<img style=\"height: 4.89ex; vertical-align: -1.55ex; \" src=\"img/img1541.svg\" alt=\"$\\displaystyle \\qquad \\cos x = \\frac{e^{ix} + e^{-ix}}{2}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n (<span class=\"arabic\">3</span>.<span class=\"arabic\">21</span>)</td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P281\" title=\"3P281\">\n Observe que o lado direito destas igualdades envolve a função exponencial de variável complexa. Precisamos definir a\n função exponencial de variável complexa e o faremos como na seção anterior onde definimos as funções trigonométricas\n de variáveis complexas. A expansão em série de potências da função <!-- MATH\n $f(x) = e^{x}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1147.svg\" alt=\"$f(x) = e^{x}$\" loading=\"lazy\"></span>, para <!-- MATH\n $x \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1151.svg\" alt=\"$x \\in \\mathbb{R}$\" loading=\"lazy\"></span>, é\n <!-- MATH\n \\begin{displaymath}\n e^{x} = 1 + x + \\frac{1}{2!} x^{2} + \\frac{1}{3!} x^{3} + \\frac{1}{4!} x^{4} + \\frac{1}{5!} x^{5} + \\frac{1}{6!} x^{6} + \\cdots.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P282\" title=\"3P282\">\n <img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1542.svg\" alt=\"$\\displaystyle e^{x} = 1 + x + \\frac{1}{2!} x^{2} + \\frac{1}{3!} x^{3} + \\frac{1}{4!} x^{4} + \\frac{1}{5!} x^{5} + \\frac{1}{6!} x^{6} + \\cdots. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P283\" title=\"3P283\">\n Mas a série de potências do lado direito da igualdade faz sentido se <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img69.svg\" alt=\"$x$\" loading=\"lazy\"></span> for um número complexo que torne a série\n convergente. Definimos então por esta série de potências a função exponencial de variável complexa dada por\n <!-- MATH\n \\begin{displaymath}\n e^{z} = 1 + z + \\frac{1}{2!} z^{2} + \\frac{1}{3!} z^{3} + \\frac{1}{4!} z^{4} + \\frac{1}{5!} z^{5} + \\frac{1}{6!} z^{6} + \\cdots,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P284\" title=\"3P284\">\n <img style=\"height: 4.56ex; vertical-align: -1.59ex; \" src=\"img/img1543.svg\" alt=\"$\\displaystyle e^{z} = 1 + z + \\frac{1}{2!} z^{2} + \\frac{1}{3!} z^{3} + \\frac{1}{4!} z^{4} + \\frac{1}{5!} z^{5} + \\frac{1}{6!} z^{6} + \\cdots, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P285\" style=\"text-indent: 0 !important;\" title=\"3P285\">\n para todo <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span> tal que a série seja convergente.\n </p>\n <p class=\" unidade\" id=\"3P286\" title=\"3P286\">\n De acordo com o teste da razão (Teorema <a href=\"#DAlembert\">3.9</a>), esta série de potências é convergente em todo o plano\n complexo, já que\n <!-- MATH\n \\begin{displaymath}\n \\lim_{n \\to \\infty} \\left| \\frac{\\frac{1}{(n+1)!}z^{n+1}}{\\frac{1}{n!}z^{n}} \\right|\n = \\lim_{n \\to \\infty} \\left| \\frac{n!}{(n+1)!} \\frac{z^{n+1}}{z^{n}} \\right| = \\lim_{n \\to \\infty} \\frac{1}{(n+1)} |z| = 0 < 1, \n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P287\" title=\"3P287\">\n <img style=\"height: 6.68ex; vertical-align: -2.74ex; \" src=\"img/img1544.svg\" alt=\"$\\displaystyle \\lim_{n \\to \\infty} \\left\\vert \\frac{\\frac{1}{(n+1)!}z^{n+1}}{\\fr...\n ...z^{n}} \\right\\vert = \\lim_{n \\to \\infty} \\frac{1}{(n+1)} \\vert z\\vert = 0 < 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P288\" style=\"text-indent: 0 !important;\" title=\"3P288\">\n para qualquer <!-- MATH\n $z \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1347.svg\" alt=\"$z \\in \\mathbb{C}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P289\" title=\"3P289\">\n Esta série é importante, porém, dificulta o trabalho com a função <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1545.svg\" alt=\"$e^{z}$\" loading=\"lazy\"></span>. Como de costume, vamos reescrever esta série\n em termos mais agradáveis. Mais precisamente, já que o lado direito da série de potências é um número complexo,\n esperamos poder escrever este número complexo na tradicional forma algébrica <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1360.svg\" alt=\"$a + bi$\" loading=\"lazy\"></span> com <!-- MATH\n $a,b \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1361.svg\" alt=\"$a, b \\in \\mathbb{R}$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P290\" title=\"3P290\">\n Tomando então <!-- MATH\n $z = x + yi$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1362.svg\" alt=\"$z = x + yi$\" loading=\"lazy\"></span> com <!-- MATH\n $x, y \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.04ex; vertical-align: -0.50ex; \" src=\"img/img1251.svg\" alt=\"$x, y \\in \\mathbb{R}$\" loading=\"lazy\"></span>, aplicando a expansão binomial, podemos reescrever a função\n exponencial na forma\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P291\" title=\"3P291\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.10ex; \" src=\"img/img1546.svg\" alt=\"$\\displaystyle e^{x + yi}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1547.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\frac{1}{n!} (x+yi)^{n}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1548.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\frac{1}{n!} \\sum_{r=0}^{n} \\frac{n!}{r!(n-...\n ...)^{r}\n = \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{r!(n-r)!} x^{n-r} (yi)^{r}.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P292\" title=\"3P292\">\n O lema a seguir nos ajudará a trabalhar com o somatório duplo do segundo membro desta última igualdade.\n \n </p>\n <div id=\"3Teo12\" title=\"3Teo12\" class=\"unidade\"><b>Lema <span class=\"arabic\">3</span>.<span class=\"arabic\">12</span></b> \n <i>Para qualquer <!-- MATH\n $m \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1369.svg\" alt=\"$m \\in \\mathbb{N}$\" loading=\"lazy\"></span>,\n </i><!-- MATH\n \\begin{displaymath}\n \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}\n = e^{x}\\frac{1}{m!}(yi)^{m} + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m+1)!(n-r)!} x^{n-r} (yi)^{r+m+1}. \n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P293\" title=\"3P293\">\n <img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1549.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r}...\n ...m_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m+1)!(n-r)!} x^{n-r} (yi)^{r+m+1}. $\" loading=\"lazy\">\n </div></div>\n \n \n <div><i>Prova</i>.\n \n Dado qualquer <!-- MATH\n $m \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1369.svg\" alt=\"$m \\in \\mathbb{N}$\" loading=\"lazy\"></span> e começando com\n <!-- MATH\n \\begin{displaymath}\n \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m},\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P294\" title=\"3P294\">\n <img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1550.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P295\" style=\"text-indent: 0 !important;\" title=\"3P295\">\n vamos separar o caso <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1375.svg\" alt=\"$n = 0$\" loading=\"lazy\"></span> do somatório externo e depois os casos <span class=\"MATH\"><img style=\"height: 1.68ex; vertical-align: -0.13ex; \" src=\"img/img1376.svg\" alt=\"$r = 0$\" loading=\"lazy\"></span> do somatório interno. Desta forma, para\n qualquer que seja <!-- MATH\n $m \\in \\mathbb{N}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1369.svg\" alt=\"$m \\in \\mathbb{N}$\" loading=\"lazy\"></span>, obtemos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P296\" title=\"3P296\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1551.svg\" alt=\"$\\displaystyle \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.04ex; vertical-align: -2.07ex; \" src=\"img/img1552.svg\" alt=\"$\\displaystyle \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 0.19ex; vertical-align: -0.10ex; \" src=\"img/img1379.svg\" alt=\"$\\displaystyle %\n $\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1553.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} x^{0} (yi)^{m} + \\sum_{n=1}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.67ex; vertical-align: -2.90ex; \" src=\"img/img1554.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} + \\sum_{n=1}^{\\infty} \\left( \\frac{1}{m!n...\n ...{n} (yi)^{m}\n + \\sum_{r=1}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m} \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.53ex; vertical-align: -2.90ex; \" src=\"img/img1555.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} + \\sum_{n=1}^{\\infty} \\frac{1}{m!n!} x^{n...\n ...\n + \\sum_{n=1}^{\\infty} \\sum_{r=1}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.67ex; vertical-align: -2.90ex; \" src=\"img/img1556.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} \\left( 1 + \\sum_{n=1}^{\\infty} \\frac{1}{n...\n ...\n + \\sum_{n=1}^{\\infty} \\sum_{r=1}^{n} \\frac{1}{(r+m)!(n-r)!} x^{n-r} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.91ex; vertical-align: -2.91ex; \" src=\"img/img1557.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} e^{x} + \\sum_{n=0}^{\\infty} \\sum_{r=1}^{n+1} \\frac{1}{(r+m)!(n+1-r)!} x^{n+1-r} (yi)^{r+m}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1558.svg\" alt=\"$\\displaystyle = \\frac{1}{m!} (yi)^{m} e^{x} + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+m+1)!(n-r)!} x^{n-r} (yi)^{r+m+1},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P297\" style=\"text-indent: 0 !important;\" title=\"3P297\">\n como desejado.\n <span style=\"float: right\"><img style=\"height: 1.59ex; vertical-align: -0.10ex; \" src=\"img/img193.svg\" alt=\"$\\qedsymbol$\" loading=\"lazy\">\n </span></p>\n </div>\n \n <p class=\" unidade\" id=\"3P298\" title=\"3P298\">\n Usando agora repetidamente este lema, temos que\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P299\" title=\"3P299\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.10ex; \" src=\"img/img1546.svg\" alt=\"$\\displaystyle e^{x + yi}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1559.svg\" alt=\"$\\displaystyle = \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{r!(n-r)!} x^{n-r} (yi)^{r}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1560.svg\" alt=\"$\\displaystyle = e^{x} + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+1)!(n-r)!} x^{n-r} (yi)^{r+1}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1561.svg\" alt=\"$\\displaystyle = e^{x} + e^{x} (yi) + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+2)!(n-r)!} x^{n-r} (yi)^{r+2}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1562.svg\" alt=\"$\\displaystyle = e^{x} + e^{x} (yi) + e^{x} \\frac{1}{2!} (yi)^{2} + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+3)!(n-r)!} x^{n-r} (yi)^{r+3}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1563.svg\" alt=\"$\\displaystyle = e^{x} + e^{x} (yi) + e^{x} \\frac{1}{2!} (yi)^{2} + e^{x} \\frac{...\n ...\n + \\sum_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+4)!(n-r)!} x^{n-r} (yi)^{r+4}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.55ex; vertical-align: -2.91ex; \" src=\"img/img1564.svg\" alt=\"$\\displaystyle = e^{x} + e^{x} (yi) + e^{x} \\frac{1}{2!} (yi)^{2} + e^{x} \\frac{...\n ...um_{n=0}^{\\infty} \\sum_{r=0}^{n} \\frac{1}{(r+5)!(n-r)!} x^{n-r} (yi)^{r+5}, %\n $\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P300\" style=\"text-indent: 0 !important;\" title=\"3P300\">\n e assim sucessivamente. Desta forma, obtemos\n <!-- MATH\n \\begin{displaymath}\n e^{x+yi} = \\sum_{n=0}^{\\infty} e^{x} \\frac{1}{n!} (yi)^{n} = e^{x} \\sum_{n=0}^{\\infty} \\frac{1}{n!} (yi)^{n},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P301\" title=\"3P301\">\n <img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1565.svg\" alt=\"$\\displaystyle e^{x+yi} = \\sum_{n=0}^{\\infty} e^{x} \\frac{1}{n!} (yi)^{n} = e^{x} \\sum_{n=0}^{\\infty} \\frac{1}{n!} (yi)^{n}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P302\" style=\"text-indent: 0 !important;\" title=\"3P302\">\n e usando o fato de que quando <span class=\"MATH\"><img style=\"height: 1.65ex; vertical-align: -0.10ex; \" src=\"img/img1566.svg\" alt=\"$n = 2r$\" loading=\"lazy\"></span> é par, temos que\n <!-- MATH\n \\begin{displaymath}\n (yi)^{n} = (yi)^{2r} = y^{2r} i^{2r} = (-1)^{r} y^{2r},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P303\" title=\"3P303\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1567.svg\" alt=\"$\\displaystyle (yi)^{n} = (yi)^{2r} = y^{2r} i^{2r} = (-1)^{r} y^{2r}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P304\" style=\"text-indent: 0 !important;\" title=\"3P304\">\n e quando <span class=\"MATH\"><img style=\"height: 1.83ex; vertical-align: -0.27ex; \" src=\"img/img1568.svg\" alt=\"$n = 2r+1$\" loading=\"lazy\"></span> é impar,\n <!-- MATH\n \\begin{displaymath}\n (yi)^{n} = (yi)^{2r+1} = y^{2r+1} i^{2r+1} = (-1)^{r} y^{2r+1}i,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P305\" title=\"3P305\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1569.svg\" alt=\"$\\displaystyle (yi)^{n} = (yi)^{2r+1} = y^{2r+1} i^{2r+1} = (-1)^{r} y^{2r+1}i, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P306\" style=\"text-indent: 0 !important;\" title=\"3P306\">\n então podemos separar o último somatório nas suas parcelas com <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> par e com <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1118.svg\" alt=\"$n$\" loading=\"lazy\"></span> ímpar e obtemos\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P307\" title=\"3P307\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 2.13ex; vertical-align: -0.10ex; \" src=\"img/img1546.svg\" alt=\"$\\displaystyle e^{x + yi}$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.50ex; vertical-align: -2.91ex; \" src=\"img/img1570.svg\" alt=\"$\\displaystyle = e^{x} \\sum_{n=0}^{\\infty} \\frac{1}{n!} (yi)^{n}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.68ex; vertical-align: -2.91ex; \" src=\"img/img1571.svg\" alt=\"$\\displaystyle = e^{x} \\left( \\sum_{n=0}^{\\infty} \\frac{1}{(2n)!} (yi)^{2n} + \\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)!} (yi)^{2n+1} \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.68ex; vertical-align: -2.91ex; \" src=\"img/img1572.svg\" alt=\"$\\displaystyle = e^{x} \\left( \\sum_{n=0}^{\\infty} \\frac{1}{(2n)!}(-1)^{n}y^{2n} + \\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)!}(-1)^{n}y^{2n+1}i \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 6.68ex; vertical-align: -2.91ex; \" src=\"img/img1573.svg\" alt=\"$\\displaystyle = e^{x} \\left( \\sum_{n=0}^{\\infty} \\frac{1}{(2n)!}(-1)^{n}y^{2n} + i\\sum_{n=0}^{\\infty} \\frac{1}{(2n+1)!}(-1)^{n}y^{2n+1} \\right)$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td> </td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1574.svg\" alt=\"$\\displaystyle = e^{x} \\left( \\cos(y) + i{\\mathrm {sen}}(y) \\right).$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P308\" title=\"3P308\">\n Formalmente, temos então uma definição alternativa para a exponencial de um número complexo sem o uso explícito das\n séries de potência.\n </p>\n <div><b>Definição <span class=\"arabic\">3</span>.<span class=\"arabic\">13</span></b> \n Dado <!-- MATH\n $z = x + yi \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.50ex; \" src=\"img/img1575.svg\" alt=\"$z = x + yi \\in \\mathbb{C}$\" loading=\"lazy\"></span>, definimos a exponencial de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1412.svg\" alt=\"$z$\" loading=\"lazy\"></span>, como sendo o número complexo representado por\n <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1576.svg\" alt=\"$e^{x+yi}$\" loading=\"lazy\"></span> e dado por\n <!-- MATH\n \\begin{displaymath}\n e^{x+yi} = e^{x} \\left( \\cos(y) + i{\\mathrm {sen}}(y) \\right) = e^{x}\\cos(y) + ie^{x}{\\mathrm {sen}}(y).\n \\end{displaymath}\n -->\n \n <div class=\"mathdisplay unidade\" id=\"3P309\" title=\"3P309\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1577.svg\" alt=\"$\\displaystyle e^{x+yi} = e^{x} \\left( \\cos(y) + i{\\mathrm {sen}}(y) \\right) = e^{x}\\cos(y) + ie^{x}{\\mathrm {sen}}(y). $\" loading=\"lazy\">\n </div></div>\n \n <p class=\" unidade\" id=\"3P310\" title=\"3P310\">\n Agora estamos prontos para obter as identidades em (<a href=\"#idexpocirc\">3.21</a>). Dado <!-- MATH\n $u \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img57.svg\" alt=\"$u \\in \\mathbb{R}$\" loading=\"lazy\"></span>, temos desta última definição\n que\n <!-- MATH\n \\begin{displaymath}\n e^{iu} = \\cos u + i {\\mathrm {sen}}u,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P311\" title=\"3P311\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1578.svg\" alt=\"$\\displaystyle e^{iu} = \\cos u + i {\\mathrm {sen}}u, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P312\" title=\"3P312\">\n e também\n <!-- MATH\n \\begin{displaymath}\n e^{-iu} = \\cos(-u) + i{\\mathrm {sen}}(-u) = \\cos u - i{\\mathrm {sen}}u.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P313\" style=\"text-indent: 0 !important;\" title=\"3P313\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1579.svg\" alt=\"$\\displaystyle e^{-iu} = \\cos(-u) + i{\\mathrm {sen}}(-u) = \\cos u - i{\\mathrm {sen}}u. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P314\" title=\"3P314\">\n Somando estas duas últimas igualdades, temos\n <!-- MATH\n \\begin{displaymath}\n e^{iu} + e^{-iu} = 2\\cos u,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P315\" title=\"3P315\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1580.svg\" alt=\"$\\displaystyle e^{iu} + e^{-iu} = 2\\cos u, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P316\" style=\"text-indent: 0 !important;\" title=\"3P316\">\n e subtraindo a segunda da primeira, temos\n <!-- MATH\n \\begin{displaymath}\n e^{iu} - e^{-iu} = 2i{\\mathrm {sen}}u.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P317\" title=\"3P317\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1581.svg\" alt=\"$\\displaystyle e^{iu} - e^{-iu} = 2i{\\mathrm {sen}}u. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P318\" title=\"3P318\">\n Segue portanto que\n <!-- MATH\n \\begin{displaymath}\n \\cos u = \\frac{e^{iu} + e^{-iu}}{2}, \\qquad \\text{e} \\qquad {\\mathrm {sen}}u = \\frac{e^{iu} - e^{-iu}}{2i}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P319\" title=\"3P319\">\n <img style=\"height: 4.89ex; vertical-align: -1.55ex; \" src=\"img/img1582.svg\" alt=\"$\\displaystyle \\cos u = \\frac{e^{iu} + e^{-iu}}{2},$\" loading=\"lazy\"> e<img style=\"height: 4.89ex; vertical-align: -1.55ex; \" src=\"img/img1583.svg\" alt=\"$\\displaystyle \\qquad {\\mathrm {sen}}u = \\frac{e^{iu} - e^{-iu}}{2i}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P320\" title=\"3P320\">\n Estas duas igualdades são as identidades exponenciais para as funções trigonométricas circulares e são válidas para\n valores reais de <span class=\"MATH\"><img style=\"height: 1.16ex; vertical-align: -0.10ex; \" src=\"img/img1.svg\" alt=\"$u$\" loading=\"lazy\"></span>. Obviamente combinando estas duas fórmulas, podemos deduzir fórmulas exponenciais para as outras\n funções trigonométricas circulares. São elas\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P321\" title=\"3P321\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.58ex; \" src=\"img/img1584.svg\" alt=\"$\\displaystyle {\\mathrm {tg}}u$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.06ex; vertical-align: -1.72ex; \" src=\"img/img1585.svg\" alt=\"$\\displaystyle = \\frac{{\\mathrm {sen}}u}{\\cos u} = \\frac{e^{iu} - e^{-iu}}{2i} \\cdot \\frac{2}{e^{iu} + e^{-iu}} = \\frac{ie^{-iu} - ie^{iu}}{e^{iu} + e^{-iu}},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.58ex; \" src=\"img/img1586.svg\" alt=\"$\\displaystyle {\\mathrm {ctg}}u$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 5.06ex; vertical-align: -1.72ex; \" src=\"img/img1587.svg\" alt=\"$\\displaystyle = \\frac{\\cos u}{{\\mathrm {sen}}u} = \\frac{e^{iu} + e^{-iu}}{2} \\cdot \\frac{2i}{e^{iu} - e^{-iu}} = \\frac{ie^{iu} + ie^{-iu}}{e^{iu} - e^{-iu}},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1588.svg\" alt=\"$\\displaystyle \\sec u$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1589.svg\" alt=\"$\\displaystyle = \\frac{1}{\\cos u} = \\frac{2}{e^{iu} + e^{-iu}},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:right;\"><span class=\"MATH\"><img style=\"height: 1.19ex; vertical-align: -0.13ex; \" src=\"img/img1590.svg\" alt=\"$\\displaystyle \\csc u$\" loading=\"lazy\"></span></td>\n <td style=\"text-align:left;\"><span class=\"MATH\"><img style=\"height: 4.69ex; vertical-align: -1.72ex; \" src=\"img/img1591.svg\" alt=\"$\\displaystyle = \\frac{1}{{\\mathrm {sen}}u} = \\frac{2i}{e^{iu} - e^{-iu}},$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n <p class=\" unidade\" id=\"3P322\" style=\"text-indent: 0 !important;\" title=\"3P322\">\n respeitados os domínios de definição das funções.\n </p>\n:::\n\n## 3.6 Fórmulas logarítmicas para as funções trigonométricas circulares inversas {#SECTION00760000000000000000}\n\n::: {.raw_html}\n \n <p class=\" unidade\" id=\"3P323\" title=\"3P323\">\n Podemos também, como no caso hiperbólico, escrever as funções inversas das funções trigonométricas circulares em termos\n do logaritmo. Isto porque a função logaritmo de um número complexo, <span class=\"MATH\"><img style=\"height: 1.21ex; vertical-align: -0.13ex; \" src=\"img/img1592.svg\" alt=\"$\\ln z$\" loading=\"lazy\"></span> é a função inversa da exponencial <span class=\"MATH\"><img style=\"height: 1.66ex; vertical-align: -0.10ex; \" src=\"img/img1545.svg\" alt=\"$e^{z}$\" loading=\"lazy\"></span>,\n com uma certa restrição no logaritmo. Para conhecer mais sobre esta restrição, recomendamos [<a href=\"/trigonometria-hiperbolica/referencias#ZillVC\">11</a>, Zill]. Por\n hora, é suficiente saber que <!-- MATH\n $\\ln e^{(a+ib)} = (a+ib)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.64ex; vertical-align: -0.62ex; \" src=\"img/img1593.svg\" alt=\"$\\ln e^{(a+ib)} = (a+ib)$\" loading=\"lazy\"></span>, para <!-- MATH\n $a \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img284.svg\" alt=\"$a \\in \\mathbb{R}$\" loading=\"lazy\"></span> e <!-- MATH\n $b \\in (-\\pi,\\pi]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1594.svg\" alt=\"$b \\in (-\\pi,\\pi]$\" loading=\"lazy\"></span>.\n </p>\n <p class=\" unidade\" id=\"3P324\" title=\"3P324\">\n Também temos que lembrar que a conhecida “fórmula de Bháskara” continua válida para resolver equações quadráticas\n que envolvem coeficientes complexos. Mais precisamente, se <!-- MATH\n $az^{2}+bz+c = 0$\n -->\n <span class=\"MATH\"><img style=\"height: 2.20ex; vertical-align: -0.27ex; \" src=\"img/img1595.svg\" alt=\"$az^{2}+bz+c = 0$\" loading=\"lazy\"></span> é uma equação com <!-- MATH\n $a, b, c \\in \\mathbb{C}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img1596.svg\" alt=\"$a, b, c \\in \\mathbb{C}$\" loading=\"lazy\"></span> e <span class=\"MATH\"><img style=\"height: 2.06ex; vertical-align: -0.51ex; \" src=\"img/img939.svg\" alt=\"$a \\neq 0$\" loading=\"lazy\"></span>, então as soluções desta equação são dadas por\n <!-- MATH\n \\begin{displaymath}\n z = \\frac{-b + \\sqrt{b^{2} - 4ac}}{2a}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P325\" title=\"3P325\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1597.svg\" alt=\"$\\displaystyle z = \\frac{-b + \\sqrt{b^{2} - 4ac}}{2a}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P326\" title=\"3P326\">\n Note que não usamos o sinal <span class=\"MATH\"><img style=\"height: 1.60ex; vertical-align: -0.27ex; \" src=\"img/img1598.svg\" alt=\"$\\pm$\" loading=\"lazy\"></span>, porque a função raiz quadrada (potência meio) para números complexos é bivalente,\n isto é, assume dois valores, que são simétricos com relação à origem e isto substitui o sinal <span class=\"MATH\"><img style=\"height: 1.60ex; vertical-align: -0.27ex; \" src=\"img/img1598.svg\" alt=\"$\\pm$\" loading=\"lazy\"></span>. Veja\n [<a href=\"/trigonometria-hiperbolica/referencias#ZillVC\">11</a>, Zill] ou outro texto sobre números complexos para um estudo mais completo sobre raízes de um número\n complexo.\n </p>\n <p class=\" unidade\" id=\"3P327\" title=\"3P327\">\n Considerando <!-- MATH\n $w = {\\mathrm {sen}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img458.svg\" alt=\"$w = {\\mathrm {sen}}^{-1}u$\" loading=\"lazy\"></span>, para <!-- MATH\n $u \\in [-1,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img423.svg\" alt=\"$u \\in [-1,1]$\" loading=\"lazy\"></span> e <!-- MATH\n $w \\in [-\\frac{\\pi}{2},\\frac{\\pi}{2}]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img424.svg\" alt=\"$w \\in\n [-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$\" loading=\"lazy\"></span>, então temos a relação\n <!-- MATH\n \\begin{displaymath}\n u = {\\mathrm {sen}}w = \\frac{e^{iw} - e^{-iw}}{2i},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P328\" title=\"3P328\">\n <img style=\"height: 4.89ex; vertical-align: -1.55ex; \" src=\"img/img1599.svg\" alt=\"$\\displaystyle u = {\\mathrm {sen}}w = \\frac{e^{iw} - e^{-iw}}{2i}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P329\" style=\"text-indent: 0 !important;\" title=\"3P329\">\n e multiplicando esta igualdade por <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1600.svg\" alt=\"$2i e^{iw}$\" loading=\"lazy\"></span> e, organizando os termos, temos\n <!-- MATH\n \\begin{displaymath}\n (e^{iw})^{2} - 2iue^{iw} - 1 = 0.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P330\" title=\"3P330\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1601.svg\" alt=\"$\\displaystyle (e^{iw})^{2} - 2iue^{iw} - 1 = 0. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P331\" title=\"3P331\">\n Resolvendo esta equação quadrática, em termos de <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span>, segue que\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\frac{2ui + \\sqrt{-4u^{2} + 4}}{2} = ui + \\sqrt{1-u^{2}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P332\" title=\"3P332\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1603.svg\" alt=\"$\\displaystyle e^{iw} = \\frac{2ui + \\sqrt{-4u^{2} + 4}}{2} = ui + \\sqrt{1-u^{2}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P333\" title=\"3P333\">\n Como <!-- MATH\n $u \\in [-1,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img423.svg\" alt=\"$u \\in [-1,1]$\" loading=\"lazy\"></span> a raiz do segundo membro é um número real. Levando em conta que <!-- MATH\n $w \\in\n [-\\frac{\\pi}{2},\\frac{\\pi}{2}]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img424.svg\" alt=\"$w \\in\n [-\\frac{\\pi}{2}, \\frac{\\pi}{2}]$\" loading=\"lazy\"></span> então aplicando o logaritmo em ambos os membros, obtemos\n <!-- MATH\n \\begin{displaymath}\n iw = \\ln e^{iw} = \\ln(ui + \\sqrt{1-u^{2}}),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P334\" title=\"3P334\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1604.svg\" alt=\"$\\displaystyle iw = \\ln e^{iw} = \\ln(ui + \\sqrt{1-u^{2}}), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P335\" style=\"text-indent: 0 !important;\" title=\"3P335\">\n e multiplicando tudo por <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1605.svg\" alt=\"$-i$\" loading=\"lazy\"></span>, segue que\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {sen}}^{-1} u = w = -i \\ln(ui + \\sqrt{1-u^{2}}).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P336\" title=\"3P336\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1606.svg\" alt=\"$\\displaystyle {\\mathrm {sen}}^{-1} u = w = -i \\ln(ui + \\sqrt{1-u^{2}}). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P337\" title=\"3P337\">\n Para <!-- MATH\n $w = \\cos^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img467.svg\" alt=\"$w = \\cos^{-1} u$\" loading=\"lazy\"></span>, vale a relação <!-- MATH\n $u = \\cos w = \\frac{e^{iw}+e^{-iw}}{2}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.09ex; vertical-align: -0.83ex; \" src=\"img/img1607.svg\" alt=\"$u = \\cos w = \\frac{e^{iw}+e^{-iw}}{2}$\" loading=\"lazy\"></span>, com <!-- MATH\n $u \\in [-1,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img423.svg\" alt=\"$u \\in [-1,1]$\" loading=\"lazy\"></span> e <!-- MATH\n $w \\in [0,\\pi]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img564.svg\" alt=\"$w \\in [0,\\pi]$\" loading=\"lazy\"></span>.\n Procedendo como anteriormente, multiplicamos esta relação por 2 e por <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span>. Obtemos\n <!-- MATH\n \\begin{displaymath}\n 2ue^{iw} = (e^{iw})^{2} + 1,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P338\" title=\"3P338\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1608.svg\" alt=\"$\\displaystyle 2ue^{iw} = (e^{iw})^{2} + 1, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P339\" style=\"text-indent: 0 !important;\" title=\"3P339\">\n e resolvendo a equação quadrática\n <!-- MATH\n \\begin{displaymath}\n (e^{iw})^{2} - 2ue^{iw} + 1 = 0\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P340\" title=\"3P340\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1609.svg\" alt=\"$\\displaystyle (e^{iw})^{2} - 2ue^{iw} + 1 = 0 $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P341\" style=\"text-indent: 0 !important;\" title=\"3P341\">\n em <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\frac{2u + \\sqrt{4u^{2} - 4}}{2} = u + \\sqrt{u^{2} - 1}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P342\" title=\"3P342\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1610.svg\" alt=\"$\\displaystyle e^{iw} = \\frac{2u + \\sqrt{4u^{2} - 4}}{2} = u + \\sqrt{u^{2} - 1}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P343\" title=\"3P343\">\n Esta solução envolve a raiz quadrada de um número que é real e negativo, já que <!-- MATH\n $u \\in [-1,1]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img423.svg\" alt=\"$u \\in [-1,1]$\" loading=\"lazy\"></span>. Escrevemos então\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = u + i\\sqrt{1-u^{2}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P344\" title=\"3P344\">\n <img style=\"height: 2.80ex; vertical-align: -0.41ex; \" src=\"img/img1611.svg\" alt=\"$\\displaystyle e^{iw} = u + i\\sqrt{1-u^{2}}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P345\" style=\"text-indent: 0 !important;\" title=\"3P345\">\n e agora aplicando o logaritmo em ambos os membros, vem\n <!-- MATH\n \\begin{displaymath}\n iw = \\ln (u + i\\sqrt{1-u^{2}}),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P346\" title=\"3P346\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1612.svg\" alt=\"$\\displaystyle iw = \\ln (u + i\\sqrt{1-u^{2}}), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P347\" style=\"text-indent: 0 !important;\" title=\"3P347\">\n já que <!-- MATH\n $w \\in [0,\\pi]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img564.svg\" alt=\"$w \\in [0,\\pi]$\" loading=\"lazy\"></span>. Multiplicando a igualdade por <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1605.svg\" alt=\"$-i$\" loading=\"lazy\"></span> obtemos\n <!-- MATH\n \\begin{displaymath}\n \\cos^{-1} u = w = -i \\ln (u + i\\sqrt{1-u^{2}}).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P348\" title=\"3P348\">\n <img style=\"height: 3.01ex; vertical-align: -0.62ex; \" src=\"img/img1613.svg\" alt=\"$\\displaystyle \\cos^{-1} u = w = -i \\ln (u + i\\sqrt{1-u^{2}}). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P349\" title=\"3P349\">\n Considerando agora <!-- MATH\n $w = {\\mathrm {tg}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.58ex; \" src=\"img/img1614.svg\" alt=\"$w = {\\mathrm {tg}}^{-1} u$\" loading=\"lazy\"></span>, válida para todo <!-- MATH\n $u \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img57.svg\" alt=\"$u \\in \\mathbb{R}$\" loading=\"lazy\"></span> e <!-- MATH\n $w \\in (-\\pi,\\pi)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1615.svg\" alt=\"$w \\in (-\\pi,\\pi)$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n u = {\\mathrm {tg}}w = \\frac{ie^{-iw} - ie^{iw}}{e^{iw} + e^{-iw}}\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P350\" title=\"3P350\">\n <img style=\"height: 5.06ex; vertical-align: -1.72ex; \" src=\"img/img1616.svg\" alt=\"$\\displaystyle u = {\\mathrm {tg}}w = \\frac{ie^{-iw} - ie^{iw}}{e^{iw} + e^{-iw}} $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P351\" style=\"text-indent: 0 !important;\" title=\"3P351\">\n donde\n <!-- MATH\n \\begin{displaymath}\n ue^{iw} + ue^{-iw} = ie^{-iw} - ie^{iw}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P352\" title=\"3P352\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1617.svg\" alt=\"$\\displaystyle ue^{iw} + ue^{-iw} = ie^{-iw} - ie^{iw}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P353\" title=\"3P353\">\n Multiplicando por <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span> e organizando os termos temos a equação quadrática\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P354\" title=\"3P354\"><table class=\"equation\">\n <tbody><tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1618.svg\" alt=\"$\\displaystyle u(e^{iw})^{2} + u = i - i(e^{iw})^{2}$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n <tr>\n <td style=\"text-align:center;\"><span class=\"MATH\"><img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1619.svg\" alt=\"$\\displaystyle (i+u)(e^{iw})^{2} - (i-u) = 0.$\" loading=\"lazy\"></span></td>\n <td class=\"eqno\" style=\"text-align:right\">\n </td></tr>\n </tbody></table></div>\n \n \n <p class=\" unidade\" id=\"3P355\" title=\"3P355\">\n Resolvendo em <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span> obtemos,\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\sqrt{\\tfrac{i-u}{i+u}} = \\left( \\tfrac{i-u}{i+u} \\right)^{\\frac{1}{2}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P356\" title=\"3P356\">\n <img style=\"height: 4.08ex; vertical-align: -1.28ex; \" src=\"img/img1620.svg\" alt=\"$\\displaystyle e^{iw} = \\sqrt{\\tfrac{i-u}{i+u}} = \\left( \\tfrac{i-u}{i+u} \\right)^{\\frac{1}{2}}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P357\" style=\"text-indent: 0 !important;\" title=\"3P357\">\n e aplicando logaritmo em ambos os membros, já que <!-- MATH\n $w \\in (-\\pi,\\pi)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1615.svg\" alt=\"$w \\in (-\\pi,\\pi)$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n iw = \\ln (e^{iw}) = \\ln\\left( \\tfrac{i-u}{i+u} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{i-u}{i+u} \\right),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P358\" title=\"3P358\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1621.svg\" alt=\"$\\displaystyle iw = \\ln (e^{iw}) = \\ln\\left( \\tfrac{i-u}{i+u} \\right)^{\\frac{1}{2}} = \\frac{1}{2} \\ln \\left( \\tfrac{i-u}{i+u} \\right), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P359\" style=\"text-indent: 0 !important;\" title=\"3P359\">\n e multiplicando a igualdade por <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1605.svg\" alt=\"$-i$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n w = \\frac{-i}{2} \\ln \\left( \\tfrac{i-u}{i+u} \\right) = \\frac{i}{2} \\ln \\left( \\tfrac{i+u}{i-u} \\right).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P360\" title=\"3P360\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1622.svg\" alt=\"$\\displaystyle w = \\frac{-i}{2} \\ln \\left( \\tfrac{i-u}{i+u} \\right) = \\frac{i}{2} \\ln \\left( \\tfrac{i+u}{i-u} \\right). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P361\" title=\"3P361\">\n Analogamente para a cotangente inversa, temos <!-- MATH\n $w = {\\mathrm {ctg}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.58ex; \" src=\"img/img1623.svg\" alt=\"$w = {\\mathrm {ctg}}^{-1} u$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $w \\in (0,\\pi)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img565.svg\" alt=\"$w \\in (0,\\pi)$\" loading=\"lazy\"></span> e <!-- MATH\n $u \\in \\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.73ex; vertical-align: -0.18ex; \" src=\"img/img57.svg\" alt=\"$u \\in \\mathbb{R}$\" loading=\"lazy\"></span> e vale a relação\n <!-- MATH\n \\begin{displaymath}\n u = {\\mathrm {ctg}}w = \\frac{ie^{iw} + ie^{-iw}}{e^{iw} - e^{-iw}}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P362\" title=\"3P362\">\n <img style=\"height: 5.06ex; vertical-align: -1.72ex; \" src=\"img/img1624.svg\" alt=\"$\\displaystyle u = {\\mathrm {ctg}}w = \\frac{ie^{iw} + ie^{-iw}}{e^{iw} - e^{-iw}}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P363\" title=\"3P363\">\n Procedendo como no caso da tangente, chegamos a equação quadrática\n <!-- MATH\n \\begin{displaymath}\n (u-i)(e^{iw})^{2} - (u+i) = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P364\" title=\"3P364\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1625.svg\" alt=\"$\\displaystyle (u-i)(e^{iw})^{2} - (u+i) = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P365\" style=\"text-indent: 0 !important;\" title=\"3P365\">\n que resolvida nos fornece,\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\sqrt{\\tfrac{u+i}{u-i}} = \\left( \\tfrac{u+i}{u-i} \\right)^{\\frac{1}{2}},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P366\" title=\"3P366\">\n <img style=\"height: 4.08ex; vertical-align: -1.28ex; \" src=\"img/img1626.svg\" alt=\"$\\displaystyle e^{iw} = \\sqrt{\\tfrac{u+i}{u-i}} = \\left( \\tfrac{u+i}{u-i} \\right)^{\\frac{1}{2}}, $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P367\" title=\"3P367\">\n Aplicando o logaritmo e multiplicando o resultado por <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1605.svg\" alt=\"$-i$\" loading=\"lazy\"></span>, chegamos a\n <!-- MATH\n \\begin{displaymath}\n {\\mathrm {ctg}}^{-1} u = w = -i \\ln \\left( \\tfrac{u+i}{u-i} \\right)^{\\frac{1}{2}} = \\frac{-i}{2} \\ln \\left( \\tfrac{u+i}{u-i} \\right) = \\frac{i}{2} \\ln \\left( \\tfrac{u-i}{u+i} \\right) .\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P368\" title=\"3P368\">\n <img style=\"height: 4.52ex; vertical-align: -1.55ex; \" src=\"img/img1627.svg\" alt=\"$\\displaystyle {\\mathrm {ctg}}^{-1} u = w = -i \\ln \\left( \\tfrac{u+i}{u-i} \\righ...\n ...( \\tfrac{u+i}{u-i} \\right) = \\frac{i}{2} \\ln \\left( \\tfrac{u-i}{u+i} \\right) . $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P369\" title=\"3P369\">\n Considerando agora <!-- MATH\n $w = \\sec^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img1628.svg\" alt=\"$w = \\sec^{-1} u$\" loading=\"lazy\"></span>, para todo <!-- MATH\n $u \\in [1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1075.svg\" alt=\"$u \\in [1,\\infty)$\" loading=\"lazy\"></span> e <!-- MATH\n $w \\in [0,\\frac{\\pi}{2})$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img1629.svg\" alt=\"$w \\in [0,\\frac{\\pi}{2})$\" loading=\"lazy\"></span>. Tomamos a relação <!-- MATH\n $u =\n \\sec w = \\frac{2}{e^{iw} + e^{-iw}}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.05ex; vertical-align: -1.06ex; \" src=\"img/img1630.svg\" alt=\"$u =\n \\sec w = \\frac{2}{e^{iw} + e^{-iw}}$\" loading=\"lazy\"></span> e obtemos,\n <!-- MATH\n \\begin{displaymath}\n ue^{iw} + ue^{-iw} = 2.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P370\" title=\"3P370\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1631.svg\" alt=\"$\\displaystyle ue^{iw} + ue^{-iw} = 2. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P371\" title=\"3P371\">\n Multiplicando a equação por <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span> e reorganizando os termos obtemos\n <!-- MATH\n \\begin{displaymath}\n u(e^{iw})^{2} - 2e^{iw} + u = 0,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P372\" title=\"3P372\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1632.svg\" alt=\"$\\displaystyle u(e^{iw})^{2} - 2e^{iw} + u = 0, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P373\" style=\"text-indent: 0 !important;\" title=\"3P373\">\n que resolvida em <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span> nos leva a\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\frac{2 + \\sqrt{4 - 4u^{2}}}{2u} = \\frac{1+\\sqrt{1-u^{2}}}{u}.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P374\" title=\"3P374\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1633.svg\" alt=\"$\\displaystyle e^{iw} = \\frac{2 + \\sqrt{4 - 4u^{2}}}{2u} = \\frac{1+\\sqrt{1-u^{2}}}{u}. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P375\" title=\"3P375\">\n A raiz quadrada do segundo membro tem no radicando um número real negativo, já que <!-- MATH\n $u \\in [1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1075.svg\" alt=\"$u \\in [1,\\infty)$\" loading=\"lazy\"></span>. Escrevemos então\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\frac{1+i\\sqrt{u^{2}-1}}{u},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P376\" title=\"3P376\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1634.svg\" alt=\"$\\displaystyle e^{iw} = \\frac{1+i\\sqrt{u^{2}-1}}{u}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P377\" style=\"text-indent: 0 !important;\" title=\"3P377\">\n e temos\n <!-- MATH\n \\begin{displaymath}\n iw = \\ln(e^{iw}) = \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right),\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P378\" title=\"3P378\">\n <img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1635.svg\" alt=\"$\\displaystyle iw = \\ln(e^{iw}) = \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right), $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P379\" style=\"text-indent: 0 !important;\" title=\"3P379\">\n ou ainda,\n <!-- MATH\n \\begin{displaymath}\n w = -i \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P380\" title=\"3P380\">\n <img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1636.svg\" alt=\"$\\displaystyle w = -i \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P381\" title=\"3P381\">\n Finalmente, para <!-- MATH\n $w = \\csc^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img1637.svg\" alt=\"$w = \\csc^{-1} u$\" loading=\"lazy\"></span>, fazendo <!-- MATH\n $u = \\csc w = \\frac{2i}{e^{iw} - e^{-iw}}$\n -->\n <span class=\"MATH\"><img style=\"height: 3.05ex; vertical-align: -1.06ex; \" src=\"img/img1638.svg\" alt=\"$u = \\csc w = \\frac{2i}{e^{iw} - e^{-iw}}$\" loading=\"lazy\"></span> para todo <!-- MATH\n $u \\in [1, \\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img1075.svg\" alt=\"$u \\in [1,\\infty)$\" loading=\"lazy\"></span> e\n <!-- MATH\n $w \\in (0, \\frac{\\pi}{2}]$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.83ex; \" src=\"img/img1639.svg\" alt=\"$w \\in (0, \\frac{\\pi}{2}]$\" loading=\"lazy\"></span>. Temos então\n <!-- MATH\n \\begin{displaymath}\n ue^{iw} - ue^{-iw} = 2i,\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P382\" title=\"3P382\">\n <img style=\"height: 2.30ex; vertical-align: -0.27ex; \" src=\"img/img1640.svg\" alt=\"$\\displaystyle ue^{iw} - ue^{-iw} = 2i, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P383\" style=\"text-indent: 0 !important;\" title=\"3P383\">\n e então\n <!-- MATH\n \\begin{displaymath}\n u(e^{iw})^{2} -2ie^{iw} - u = 0.\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P384\" title=\"3P384\">\n <img style=\"height: 2.65ex; vertical-align: -0.62ex; \" src=\"img/img1641.svg\" alt=\"$\\displaystyle u(e^{iw})^{2} -2ie^{iw} - u = 0. $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P385\" title=\"3P385\">\n Resolvendo em <span class=\"MATH\"><img style=\"height: 2.02ex; vertical-align: -0.10ex; \" src=\"img/img1602.svg\" alt=\"$e^{iw}$\" loading=\"lazy\"></span>, vem\n <!-- MATH\n \\begin{displaymath}\n e^{iw} = \\frac{2i + \\sqrt{-4 + 4u^{2}}}{2u} = \\frac{i + \\sqrt{u^{2}-1}}{u},\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P386\" title=\"3P386\">\n <img style=\"height: 5.04ex; vertical-align: -1.55ex; \" src=\"img/img1642.svg\" alt=\"$\\displaystyle e^{iw} = \\frac{2i + \\sqrt{-4 + 4u^{2}}}{2u} = \\frac{i + \\sqrt{u^{2}-1}}{u}, $\" loading=\"lazy\">\n </div><p class=\" unidade\" id=\"3P387\" style=\"text-indent: 0 !important;\" title=\"3P387\">\n e aplicando o logaritmo e uma multiplicação por <span class=\"MATH\"><img style=\"height: 1.82ex; vertical-align: -0.27ex; \" src=\"img/img1605.svg\" alt=\"$-i$\" loading=\"lazy\"></span>, temos\n <!-- MATH\n \\begin{displaymath}\n \\csc^{-1} u = w = -i \\ln \\left(\\tfrac{i + \\sqrt{u^{2}-1}}{u} \\right).\n \\end{displaymath}\n -->\n </p>\n <div class=\"mathdisplay unidade\" id=\"3P388\" title=\"3P388\">\n <img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1643.svg\" alt=\"$\\displaystyle \\csc^{-1} u = w = -i \\ln \\left(\\tfrac{i + \\sqrt{u^{2}-1}}{u} \\right). $\" loading=\"lazy\">\n </div>\n \n <p class=\" unidade\" id=\"3P389\" title=\"3P389\">\n Organizando as fórmulas logarítmicas, temos a tabela abaixo.\n </p>\n \n <div class=\"CENTER\"><a id=\"5383\"></a>\n <table>\n <caption><strong>Tabela 3.2:</strong>\n Fórmulas logarítmicas para as funções trigonométricas circulares inversas.</caption>\n <tbody><tr><td>\n <div class=\"CENTER\">\n <table class=\"PAD BORDER\">\n <tbody><tr><td class=\"CENTER\">função</td>\n <td class=\"CENTER\">domínio</td>\n <td class=\"CENTER\">fórmula logarítmica</td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n ${\\mathrm {sen}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img428.svg\" alt=\"${\\mathrm {sen}}^{-1}u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img280.svg\" alt=\"$[-1,1]$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $-i \\ln(ui + \\sqrt{1-u^{2}})$\n -->\n <span class=\"MATH\"><img style=\"height: 2.70ex; vertical-align: -0.62ex; \" src=\"img/img1644.svg\" alt=\"$-i \\ln(ui + \\sqrt{1-u^{2}})$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n $\\cos^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img521.svg\" alt=\"$\\cos^{-1} u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img280.svg\" alt=\"$[-1,1]$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $-i \\ln (u + i\\sqrt{1-u^{2}})$\n -->\n <span class=\"MATH\"><img style=\"height: 2.70ex; vertical-align: -0.62ex; \" src=\"img/img1645.svg\" alt=\"$-i \\ln (u + i\\sqrt{1-u^{2}})$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n ${\\mathrm {tg}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.58ex; \" src=\"img/img522.svg\" alt=\"${\\mathrm {tg}}^{-1} u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <!-- MATH\n $\\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.64ex; vertical-align: -0.10ex; \" src=\"img/img217.svg\" alt=\"$\\mathbb{R}$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $\\frac{i}{2} \\ln \\left( \\tfrac{i+u}{i-u} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.94ex; vertical-align: -0.95ex; \" src=\"img/img1646.svg\" alt=\"$\\frac{i}{2} \\ln \\left( \\tfrac{i+u}{i-u} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n ${\\mathrm {ctg}}^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.50ex; vertical-align: -0.58ex; \" src=\"img/img523.svg\" alt=\"${\\mathrm {ctg}}^{-1} u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <!-- MATH\n $\\mathbb{R}$\n -->\n <span class=\"MATH\"><img style=\"height: 1.64ex; vertical-align: -0.10ex; \" src=\"img/img217.svg\" alt=\"$\\mathbb{R}$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $\\frac{i}{2} \\ln \\left( \\tfrac{u-i}{u+i} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.94ex; vertical-align: -0.95ex; \" src=\"img/img1647.svg\" alt=\"$\\frac{i}{2} \\ln \\left( \\tfrac{u-i}{u+i} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n $\\sec^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img524.svg\" alt=\"$\\sec^{-1} u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <!-- MATH\n $[1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img547.svg\" alt=\"$[1,\\infty)$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $-i \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1648.svg\" alt=\"$-i \\ln \\left( \\tfrac{1+i\\sqrt{u^{2}-1}}{u} \\right)$\" loading=\"lazy\"></span> <br>\n <br></td>\n </tr>\n <tr><td class=\"CENTER\"><!-- MATH\n $\\csc^{-1} u$\n -->\n <span class=\"MATH\"><img style=\"height: 2.05ex; vertical-align: -0.13ex; \" src=\"img/img525.svg\" alt=\"$\\csc^{-1} u$\" loading=\"lazy\"></span></td>\n <td class=\"CENTER\"> <!-- MATH\n $[1,\\infty)$\n -->\n <span class=\"MATH\"><img style=\"height: 2.30ex; vertical-align: -0.62ex; \" src=\"img/img547.svg\" alt=\"$[1,\\infty)$\" loading=\"lazy\"></span> </td>\n <td class=\"CENTER\"><!-- MATH\n $-i \\ln \\left(\\tfrac{i + \\sqrt{u^{2}-1}}{u} \\right)$\n -->\n <span class=\"MATH\"><img style=\"height: 3.98ex; vertical-align: -1.47ex; \" src=\"img/img1649.svg\" alt=\"$-i \\ln \\left(\\tfrac{i + \\sqrt{u^{2}-1}}{u} \\right)$\" loading=\"lazy\"></span></td>\n </tr>\n </tbody></table>\n </div></td></tr>\n </tbody></table>\n </div>\n \n::: \n\n```{=html}\n\n</div>\n\n```","srcMarkdownNoYaml":""},"formats":{"moan-livro-html":{"identifier":{"display-name":"HTML","target-format":"moan-livro-html","base-format":"html","extension-name":"moan-livro"},"execute":{"fig-width":7,"fig-height":5,"fig-format":"retina","fig-dpi":96,"df-print":"default","error":false,"eval":true,"cache":null,"freeze":false,"echo":true,"output":true,"warning":true,"include":true,"keep-md":false,"keep-ipynb":false,"ipynb":null,"enabled":null,"daemon":null,"daemon-restart":false,"debug":false,"ipynb-filters":[],"ipynb-shell-interactivity":null,"plotly-connected":true,"engine":"markdown"},"render":{"keep-tex":false,"keep-typ":false,"keep-source":false,"keep-hidden":false,"prefer-html":false,"output-divs":true,"output-ext":"html","fig-align":"default","fig-pos":null,"fig-env":null,"code-fold":"none","code-overflow":"scroll","code-link":false,"code-line-numbers":false,"code-tools":false,"tbl-colwidths":"auto","merge-includes":true,"inline-includes":false,"preserve-yaml":false,"latex-auto-mk":true,"latex-auto-install":true,"latex-clean":true,"latex-min-runs":1,"latex-max-runs":10,"latex-makeindex":"makeindex","latex-makeindex-opts":[],"latex-tlmgr-opts":[],"latex-input-paths":[],"latex-output-dir":null,"link-external-icon":false,"link-external-newwindow":false,"self-contained-math":false,"format-resources":[],"notebook-links":true,"shortcodes":[],"format-links":false},"pandoc":{"standalone":true,"wrap":"none","default-image-extension":"png","to":"html","filters":["lightbox"],"include-after-body":{"text":"<script 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Pro","citation":true,"google-scholar":true,"smooth-scroll":true,"theme":{"light":["cosmo","_extensions/moan-livro/custom.scss","estilos.css"],"dark":["superhero","_extensions/moan-livro/custom.scss","estilos.css"]},"revealjs-plugins":[],"lightbox":"auto","crossref":{"chapters":true},"controle-moan":{"dicionario":false},"moan-dados":{"ark":"68745/eMR8J","título":"As funções trigonométricas circulares e hiperbólicas","formato":"Livro Digital (online)","autor":"Sandro Marcos Guzzo","apontamento":"https://livro.online/trigonometria-hiperbolica","edição":1,"descrição":"Esta obra trata da construção da trigonometria hiperbólica na hipérbole trigonométrica, fazendo a comparação com a trigonometria circular. São abordados aspectos das funções trigonométricas circulares e hiperbólicas, relacionados ao cálculo diferencial e integral.","editora":"Editora Moan","local":"Foz do Iguaçu - PR, Brasil","editor":"Rafael Tavares Juliani","diagramador":"Rafael Tavares Juliani","capista":"Rafael Tavares Juliani","revisor":"Sandra Maria Tieppo e Emerson Mário Boldo","palavras-chave":"trigonometria,trigonometria hiperbólica,funções trigonométricas,matemática","área":"Matemática","cdd":516.24,"bisac":"MAT012000 e MAT032000","ano-publicação":2021,"data-publicação":"22/03/2021","idioma":"Português do Brasil (pt-BR)","país":"Brasil","faixa-etária":"Acima de 12 anos","ark-versao-impressa":"68745/eMR8J.4N","isbn-versao-impressa":9786599140440,"direitos-autorais":"© Sandro Marcos Guzzo e a Editora Moan, 2021","licenca":"Todos os direitos reservados","licenca-link":"https://www.planalto.gov.br/ccivil_03/leis/l9610.htm"},"bibliography":["referencias.bib"],"csl":"_extensions/moan-livro/universidade-estadual-de-alagoas-abnt.csl","_quarto-vars":{"e-mail":"[editora@livro.online](mailto:editora@livro.online)","whatsapp":"[+55 (45) 9 3505-0721](https://api.whatsapp.com/send?phone=5545935050721)"},"comments":{"hypothesis":{"theme":"clean","openSidebar":false}}},"extensions":{"book":{"multiFile":true}}}},"projectFormats":["moan-livro-html"]} |