{"id":32690,"date":"2019-11-07T14:38:33","date_gmt":"2019-11-07T17:38:33","guid":{"rendered":"https:\/\/blog.estrategiavestibulares.com.br\/?p=32690"},"modified":"2023-06-23T16:05:36","modified_gmt":"2023-06-23T19:05:36","slug":"formula-de-de-moivre","status":"publish","type":"post","link":"https:\/\/vestibulares.estrategia.com\/portal\/materias\/matematica\/formula-de-de-moivre\/","title":{"rendered":"F\u00f3rmula de De Moivre: conceito e como \u00e9 cobrado em prova?"},"content":{"rendered":"<p>A <strong>F&oacute;rmula de De Moivre<\/strong> &eacute; utilizada para se realizar opera&ccedil;&otilde;es de potencia&ccedil;&atilde;o e de radicia&ccedil;&atilde;o com <strong>n&uacute;meros complexos<\/strong>.<p><!--[CDATA[\n\n<p--><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bn%7D=%7Cz%7C%5E%7Bn%7Dcis(n%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large \\dpi{150} \\large z^{n}=|z|^{n}cis(n\\theta )\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" class=\"\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Csqrt%5Bn%5D%7B%7Cz%7C%7Dcis(%5Cfrac%7B%5Ctheta&amp;space;+2k%5Cpi%7D%7Bn%7D),k=0&amp;space;,&amp;space;1,&amp;space;...,&amp;space;n-1\" alt=\"\\dpi{150} \\large \\sqrt[n]{|z|}cis(\\frac{\\theta +2k\\pi}{n}),k=0 , 1, ..., n-1\" width=\"365\" height=\"50\" align=\"absmiddle\"><\/p><p>Vamos ver alguns <strong>conceitos iniciais<\/strong> necess&aacute;rios para entender as f&oacute;rmulas:<\/p><p>Um <strong>n&uacute;mero complexo<\/strong> &eacute; um n&uacute;mero de forma <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=a+bi\" alt=\"\\dpi{150} \\large z=a+bi\" align=\"absmiddle\">, em que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;a,&amp;space;b&amp;space;%5Cin&amp;space;%5Cmathbb%7BR%7D\" alt=\"\\dpi{150} \\large a, b \\in \\mathbb{R}\" align=\"absmiddle\">&nbsp; e&nbsp; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;i%5E%7B2%7D=-1\" alt=\"\\dpi{150} \\large i^{2}=-1\" align=\"absmiddle\"> &eacute; a unidade imagin&aacute;ria. Esse n&uacute;mero, se escrito dessa forma est&aacute; em sua <strong>forma alg&eacute;brica<\/strong>.<\/p><p>Outra forma de se interpretar os <a href=\"https:\/\/vestibulares.estrategia.com\/portal\/matematica\/historia-dos-numeros-complexos\/\" target=\"_blank\" rel=\"noopener\">n&uacute;meros complexos<\/a> &eacute; exager&aacute;-los como pares ordenados em um plano, o chamado <strong>Plano de Argand-Gauss<\/strong>. Nesse plano, colocamos a parte real dos n&uacute;meros complexos no eixo das abscissas e a parte imagin&aacute;ria no eixo das ordenadas.<\/p><p><img decoding=\"async\" class=\"alignnone size-medium wp-image-32698\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/11\/formula-de-de-moivre.jpg\" alt=\"formula de de moivre\" width=\"300\" height=\"252\"><\/p><p>Interpretando os n&uacute;meros complexos dessa forma, podemos introduzir a forma trigonom&eacute;trica dos n&uacute;meros complexos, em que eles s&atilde;o dados por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%7Cz%7C(cos%5Ctheta&amp;space;+isen%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z=|z|(cos\\theta +isen\\theta )\" align=\"absmiddle\">, e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;(cos%5Ctheta&amp;space;+&amp;space;isen%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large (cos\\theta + isen\\theta )\" align=\"absmiddle\"> pode ser abreviado por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cis%5Ctheta\" alt=\"\\dpi{150} \\large cis\\theta\" align=\"absmiddle\">, ou seja, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%7Cz%7C(cos%5Ctheta&amp;space;+isen%5Ctheta&amp;space;)=%7Cz%7Ccis%5Ctheta\" alt=\"\\dpi{150} \\large z=|z|(cos\\theta +isen\\theta )=|z|cis\\theta\" align=\"absmiddle\">.<\/p><p>Nessa express&atilde;o, temos que, se o n&uacute;mero na forma alg&eacute;brica &eacute; dado por <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=a+bi\" alt=\"\\dpi{150} \\large z=a+bi\" align=\"absmiddle\">, ent&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%7Cz%7C=%5Csqrt%7Ba%5E%7B2%7D+b%5E%7B2%7D%7D\" alt=\"\\dpi{150} \\large |z|=\\sqrt{a^{2}+b^{2}}\" align=\"absmiddle\"> &eacute; o m&oacute;dulo de z, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cos%5Ctheta&amp;space;=%5Cfrac%7Ba%7D%7B%5Csqrt%7Ba%5E%7B2%7D+b%5E%7B2%7D%7D%7D\" alt=\"\\dpi{150} \\large cos\\theta =\\frac{a}{\\sqrt{a^{2}+b^{2}}}\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;sen%5Ctheta&amp;space;=%5Cfrac%7Bb%7D%7B%5Csqrt%7Ba%5E%7B2%7D+b%5E%7B2%7D%7D%7D\" alt=\"\\dpi{150} \\large sen\\theta =\\frac{b}{\\sqrt{a^{2}+b^{2}}}\" align=\"absmiddle\">, em que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Ctheta%5Cin&amp;space;%5B0,2%5Cpi&amp;space;%5D\" alt=\"\\dpi{150} \\large \\theta\\in [0,2\\pi ]\" align=\"absmiddle\">) &eacute; o <strong>argumento de z<\/strong>.<\/p><p>As f&oacute;rmulas de De Moivre utilizam a forma trigonom&eacute;trica dos n&uacute;meros complexos para realizar opera&ccedil;&otilde;es de potencia&ccedil;&atilde;o e radicia&ccedil;&atilde;o.<\/p><div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-transparent ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\"><p class=\"ez-toc-title\" style=\"cursor:inherit\">Navegue pelo conte\u00fado<\/p>\n<\/div><nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/matematica\/formula-de-de-moivre\/#Primeira-Formula-de-De-Moivre\" >Primeira F&oacute;rmula de De Moivre<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/matematica\/formula-de-de-moivre\/#Segunda-Formula-de-De-Moivre\" >Segunda F&oacute;rmula de De Moivre<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/vestibulares.estrategia.com\/portal\/materias\/matematica\/formula-de-de-moivre\/#Questao-de-Vestibular\" >Quest&atilde;o de Vestibular<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\" id=\"h-primeira-formula-de-de-moivre\"><span class=\"ez-toc-section\" id=\"Primeira-Formula-de-De-Moivre\"><\/span>Primeira F&oacute;rmula de De Moivre<span class=\"ez-toc-section-end\"><\/span><\/h2><h3 class=\"wp-block-heading\" id=\"h-potenciacao\"><strong>Potencia&ccedil;&atilde;o<\/strong><\/h3><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bn%7D=%7Cz%7C%5E%7Bn%7Dcis(n%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z^{n}=|z|^{n}cis(n\\theta )\" align=\"absmiddle\"><\/p><p>Utiliza-se a primeira f&oacute;rmula de De Moivre para calcular a n-&eacute;sima pot&ecirc;ncia de um n&uacute;mero complexo z. Para isso, basta elevar seu m&oacute;dulo &agrave; n-&eacute;sima pot&ecirc;ncia e multiplicar seu argumento por n.<\/p><h2 class=\"wp-block-heading\" id=\"h-segunda-formula-de-de-moivre\"><span class=\"ez-toc-section\" id=\"Segunda-Formula-de-De-Moivre\"><\/span>Segunda F&oacute;rmula de De Moivre<span class=\"ez-toc-section-end\"><\/span><\/h2><h3 class=\"wp-block-heading\" id=\"h-radiciacao\"><strong>Radicia&ccedil;&atilde;o<\/strong><\/h3><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Csqrt%5Bn%5D%7Bz%7D=%5Csqrt%5Bn%5D%7B%7Cz%7C%7Dcis&amp;space;(%5Cfrac%7B%5Ctheta&amp;space;+2k%5Cpi&amp;space;%7D%7Bn%7D),k=0,1,...,n-1\" alt=\"\\dpi{150} \\large \\sqrt[n]{z}=\\sqrt[n]{|z|}cis (\\frac{\\theta +2k\\pi }{n}),k=0,1,...,n-1\" align=\"absmiddle\"><\/p><p>A segunda f&oacute;rmula de De Moivre nos diz que um n&uacute;mero complexo z possui n ra&iacute;zes n-&eacute;simas, dadas pela substitui&ccedil;&atilde;o dos valores de k por 0, 1&hellip; n-1.<\/p><h2 class=\"wp-block-heading\" id=\"h-questao-de-vestibular\"><span class=\"ez-toc-section\" id=\"Questao-de-Vestibular\"><\/span>Quest&atilde;o de Vestibular<span class=\"ez-toc-section-end\"><\/span><\/h2><h3 class=\"wp-block-heading\" id=\"h-ita-2018\"><strong><a href=\"https:\/\/vestibulares.estrategia.com\/portal\/vestibulares\/vestibular-ita\/\" target=\"_blank\">ITA<\/a><\/strong> <strong>2018<\/strong><\/h3><p>As ra&iacute;zes do polin&ocirc;mio <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;1+z+z%5E%7B2%7D+z%5E%7B3%7D+z%5E%7B4%7D+z%5E%7B5%7D+z%5E%7B6%7D+z%5E%7B7%7D\" alt=\"\\dpi{150} \\large 1+z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}+z^{7}\" align=\"absmiddle\">, quando representados no plano complexo, formam os v&eacute;rtices de um pol&iacute;gono complexo cuja &aacute;rea &eacute;:<\/p><p>a) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7B%5Csqrt%7B2%7D-1%7D%7B2%7D\" alt=\"\\dpi{150} \\large \\frac{\\sqrt{2}-1}{2}\" align=\"absmiddle\"><\/p><p>b) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7B%5Csqrt%7B2%7D+1%7D%7B2%7D\" alt=\"\\dpi{150} \\large \\frac{\\sqrt{2}+1}{2}\" align=\"absmiddle\"><\/p><p>c) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Csqrt%7B2%7D\" alt=\"\\dpi{150} \\large \\sqrt{2}\" align=\"absmiddle\"><\/p><p>d) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7B3%5Csqrt%7B2%7D+1%7D%7B2%7D\" alt=\"\\dpi{150} \\large \\frac{3\\sqrt{2}+1}{2}\" align=\"absmiddle\"><\/p><p>e) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;3%5Csqrt%7B2%7D\" alt=\"\\dpi{150} \\large 3\\sqrt{2}\" align=\"absmiddle\"><\/p><h3><strong>Resolu&ccedil;&atilde;o comentada<\/strong><\/h3><p>Vamos, inicialmente, analisar o polin&ocirc;mio dado, para encontrarmos as ra&iacute;zes. Nota-se que, para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5Cneq&amp;space;1\" alt=\"\\dpi{150} \\large z\\neq 1\" align=\"absmiddle\">, o polin&ocirc;mio pode ser visto como uma PG de raz&atilde;o z. Ent&atilde;o podemos aplicar a f&oacute;rmula de soma de PG para a soma <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;1+z+z%5E%7B2%7D+z%5E%7B3%7D+z%5E%7B4%7D+z%5E%7B5%7D+z%5E%7B6%7D+z%5E%7B7%7D\" alt=\"\\dpi{150} \\large 1+z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}+z^{7}\" align=\"absmiddle\">, e obtemos <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7Bz%5E%7B2%7D-1%7D%7Bz-1%7D\" alt=\"\\dpi{150} \\large \\frac{z^{2}-1}{z-1}\" align=\"absmiddle\">. Queremos os valores de z para os quais essa express&atilde;o &eacute; nula, e para isso basta que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7B8%7D-1=0%5Crightarrow&amp;space;z%5E%7B8%7D=1\" alt=\"\\dpi{150} \\large z^{8}-1=0\\rightarrow z^{8}=1\" align=\"absmiddle\">.<\/p><p>Assim, vemos que as ra&iacute;zes do polin&ocirc;mio s&atilde;o as ra&iacute;zes oitavas de 1, ou seja, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%5Csqrt%5B8%5D%7B1%7D\" alt=\"\\dpi{150} \\large z=\\sqrt[8]{1}\" align=\"absmiddle\">.<\/p><p>Vamos aplicar a segunda f&oacute;rmula de De Moivre, sabendo que o m&oacute;dulo do n&uacute;mero real 1 &eacute; 1 e seu argumento &eacute; 0:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%5Csqrt%5B8%5D%7B1%7D%5Crightarrow&amp;space;z=cis(%5Cfrac%7B0+2k%5Cpi&amp;space;%7D%7B8%7D)%5Crightarrow&amp;space;z=cis(%5Cfrac%7Bk%5Cpi&amp;space;%7D%7B4%7D)\" alt=\"\\dpi{150} \\large z=\\sqrt[8]{1}\\rightarrow z=cis(\\frac{0+2k\\pi }{8})\\rightarrow z=cis(\\frac{k\\pi }{4})\" align=\"absmiddle\">, com <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;k=0,1,...,7\" alt=\"\\dpi{150} \\large k=0,1,...,7\" align=\"absmiddle\">.<\/p><p><strong>Lembrando:<\/strong> supomos inicialmente que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5Cneq&amp;space;1\" alt=\"\\dpi{150} \\large z\\neq 1\" align=\"absmiddle\">, o que exclui o caso <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;k=0\" alt=\"\\dpi{150} \\large k=0\" align=\"absmiddle\">.<\/p><p>Sendo assim, as ra&iacute;zes do polin&ocirc;mio s&atilde;o <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cis(%5Cfrac%7B%5Cpi&amp;space;%7D%7B4%7D),&amp;space;cis(%5Cfrac%7B2%5Cpi&amp;space;%7D%7B4%7D),...,cis(%5Cfrac%7B7%5Cpi&amp;space;%7D%7B4%7D)\" alt=\"\\dpi{150} \\large cis(\\frac{\\pi }{4}), cis(\\frac{2\\pi }{4}),...,cis(\\frac{7\\pi }{4})\" align=\"absmiddle\">.<\/p><p>No plano complexo, esses 7 n&uacute;meros representam 7 pontos sobre uma circunfer&ecirc;ncia na origem, e est&atilde;o separados por &acirc;ngulos de <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7B%5Cpi&amp;space;%7D%7B4%7D\" alt=\"\\dpi{150} \\large \\frac{\\pi }{4}\" align=\"absmiddle\">, ou seja:<\/p><p><img decoding=\"async\" class=\"alignnone size-medium wp-image-32702\" src=\"https:\/\/cdn.blog.estrategiavestibulares.com.br\/vestibulares\/wp-content\/uploads\/2019\/11\/CapturarDe-Moivre.png\" alt=\"De Moivre\" width=\"300\" height=\"253\"><\/p><p>Para calcular a &aacute;rea desse pol&iacute;gono, basta dividir o pol&iacute;gono: Temos que o<br>centro da circunfer&ecirc;ncia e os pontos 1 e 2 formam um tri&acirc;ngulo de lados 1 e<br>&acirc;ngulo central <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5Cfrac%7B%5Cpi&amp;space;%7D%7B4%7D\" alt=\"\\dpi{150} \\large \\frac{\\pi }{4}\" align=\"absmiddle\">. Assim, podemos calcular a &aacute;rea desse tri&acirc;ngulo como<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;S_%7B1%7D=%5Cfrac%7B1%7D%7B2%7D%5Cast&amp;space;1%5Cast&amp;space;1%5Cast&amp;space;sen(%5Cfrac%7B%5Cpi&amp;space;%7D%7B4%7D)=%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D\" alt=\"\\dpi{150} \\large S_{1}=\\frac{1}{2}\\ast 1\\ast 1\\ast sen(\\frac{\\pi }{4})=\\frac{\\sqrt{2}}{4}\" align=\"absmiddle\">. Podemos formar tri&acirc;ngulos id&ecirc;nticos a esse com os pontos 2 e 3, 3 e 4, 4 e 5, 5 e 6, 6 e 7, ressaltando em 6 tri&acirc;ngulos com o mesmo valor de &aacute;rea.<\/p><p>Resta, ainda, o tri&acirc;ngulo formado pelos pontos 1 e 7, que formam um tri&acirc;ngulo ret&acirc;ngulo. Ent&atilde;o, sua &aacute;rea &eacute; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;S_%7B2%7D=%5Cfrac%7B1%7D%7B2%7D%5Cast&amp;space;1%5Cast&amp;space;1=%5Cfrac%7B1%7D%7B2%7D\" alt=\"\\dpi{150} \\large S_{2}=\\frac{1}{2}\\ast 1\\ast 1=\\frac{1}{2}\" align=\"absmiddle\">.<\/p><p>A &aacute;rea total do pol&iacute;gono ser&aacute;, ent&atilde;o, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;S_%7Bt%7D=6S_%7B1%7D+S_%7B2%7D=6%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D+%5Cfrac%7B1%7D%7B2%7D=%5Cfrac%7B3%5Csqrt%7B2%7D+1%7D%7B2%7D\" alt=\"\\dpi{150} \\large S_{t}=6S_{1}+S_{2}=6\\frac{\\sqrt{2}}{4}+\\frac{1}{2}=\\frac{3\\sqrt{2}+1}{2}\" align=\"absmiddle\">.<\/p><h3><strong>Demonstra&ccedil;&atilde;o<\/strong><\/h3><p>Para demonstrar a primeira f&oacute;rmula de De Moire, usaremos uma indu&ccedil;&atilde;o. Seja <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%7Cz%7C(cos%5Ctheta&amp;space;+isen%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z=|z|(cos\\theta +isen\\theta )\" align=\"absmiddle\">. Vamos analisar o que ocorre para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7B2%7D\" alt=\"\\dpi{150} \\large z^{2}\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7B2%7D=%7Cz%7C%5E%7B2%7D(cos%5Ctheta&amp;space;+&amp;space;isen%5Ctheta&amp;space;)%5E%7B2%7D=%7Cz%7C%5E%7B2%7D%5Bcos%5E%7B2%7D%5Ctheta&amp;space;+2cos%5Ctheta&amp;space;%5Cast&amp;space;isen%5Ctheta+(isen%5Ctheta)%5E%7B2%7D%5D\" alt=\"\\dpi{150} \\large z^{2}=|z|^{2}(cos\\theta + isen\\theta )^{2}=|z|^{2}[cos^{2}\\theta +2cos\\theta \\ast isen\\theta+(isen\\theta)^{2}]\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7B2%7D=%7Cz%7C%5E%7B2%7D%5B(cos%5E%7B2%7D%5Ctheta&amp;space;-sen%5E%7B2%7D%5Ctheta&amp;space;)%5D+i%5Cast&amp;space;(2sen%5Ctheta&amp;space;cos%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z^{2}=|z|^{2}[(cos^{2}\\theta -sen^{2}\\theta )]+i\\ast (2sen\\theta cos\\theta )\" align=\"absmiddle\"><\/p><p>Das f&oacute;rmulas de arcos duplos, sabemos que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;(cos%5E%7B2%7D%5Ctheta&amp;space;-sen%5E%7B2%7D%5Ctheta&amp;space;)=cos(2%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large (cos^{2}\\theta -sen^{2}\\theta )=cos(2\\theta )\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;2sen%5Ctheta&amp;space;cos%5Ctheta&amp;space;=&amp;space;sen(2%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large 2sen\\theta cos\\theta = sen(2\\theta )\" align=\"absmiddle\">. Assim, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7B2%7D=%7Cz%7C%5E%7B2%7Dcis(2%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z^{2}=|z|^{2}cis(2\\theta )\" align=\"absmiddle\">.<\/p><p>Suponha que, para um certo k, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bk%7D=%7Cz%7C%5E%7Bk%7Dcis(k%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z^{k}=|z|^{k}cis(k\\theta )\" align=\"absmiddle\">. Analisemos <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bk%7D%5Cast&amp;space;z=z%5E%7Bk+1%7D\" alt=\"\\dpi{150} \\large z^{k}\\ast z=z^{k+1}\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bk%7D*z=%7Cz%7C%5E%7Bk%7D(cos(k%5Ctheta&amp;space;)+isen(k%5Ctheta&amp;space;))*%7Cz%7C(cos%5Ctheta&amp;space;+isen%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z^{k}*z=|z|^{k}(cos(k\\theta )+isen(k\\theta ))*|z|(cos\\theta +isen\\theta )\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bk+1%7D=%7Cz%7C%5E%7Bk+1%7D%5Bcos(k%5Ctheta&amp;space;)cos%5Ctheta&amp;space;+i%5E%7B2%7Dsen(k%5Ctheta&amp;space;)sen%5Ctheta&amp;space;+i(sen(k%5Ctheta&amp;space;)cos%5Ctheta&amp;space;+sen%5Ctheta&amp;space;cos(k%5Ctheta))%5D\" alt=\"\\dpi{150} \\large z^{k+1}=|z|^{k+1}[cos(k\\theta )cos\\theta +i^{2}sen(k\\theta )sen\\theta +i(sen(k\\theta )cos\\theta +sen\\theta cos(k\\theta))]\" align=\"absmiddle\"><\/p><p>Das f&oacute;rmulas de soma de arcos:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cosacosb&amp;space;-senasenb=cos(a+b)\" alt=\"\\dpi{150} \\large cosacosb -senasenb=cos(a+b)\" align=\"absmiddle\"> e&nbsp;<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;senacosb&amp;space;+&amp;space;senbcoasa=sen(a+b)\" alt=\"\\dpi{150} \\large senacosb + senbcoasa=sen(a+b)\" align=\"absmiddle\"> vemos que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cos&amp;space;(k%5Ctheta&amp;space;)cos%5Ctheta&amp;space;+i%5E%7B2%7Dsen(k%5Ctheta&amp;space;)sen%5Ctheta&amp;space;=cos&amp;space;(k%5Ctheta&amp;space;)cos%5Ctheta&amp;space;-&amp;space;sen(k%5Ctheta&amp;space;)sen%5Ctheta=cos%5B(k+1)%5Ctheta&amp;space;%5D\" alt=\"\\dpi{150} \\large cos (k\\theta )cos\\theta +i^{2}sen(k\\theta )sen\\theta =cos (k\\theta )cos\\theta - sen(k\\theta )sen\\theta=cos[(k+1)\\theta ]\" align=\"absmiddle\"><\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;i%5Bsen(k%5Ctheta&amp;space;)cos%5Ctheta&amp;space;+sen%5Ctheta&amp;space;cos(k%5Ctheta&amp;space;)%5D=isen%5B(k+1)%5Ctheta&amp;space;%5D\" alt=\"\\dpi{150} \\large i[sen(k\\theta )cos\\theta +sen\\theta cos(k\\theta )]=isen[(k+1)\\theta ]\" align=\"absmiddle\">. Ent&atilde;o: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z%5E%7Bk+1%7D=%7Cz%7C%5E%7Bk+1%7Dcis%5B(k+1)%5Ctheta&amp;space;%5D\" alt=\"\\dpi{150} \\large z^{k+1}=|z|^{k+1}cis[(k+1)\\theta ]\" align=\"absmiddle\">, o que conclui a demonstra&ccedil;&atilde;o.<\/p><p>Tendo demonstrado a primeira f&oacute;rmula de De Moivre, podemos us&aacute;-la para demonstrar a segunda:<\/p><p>Suponha que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;z=%7Cz%7Ccis(%5Ctheta&amp;space;)\" alt=\"\\dpi{150} \\large z=|z|cis(\\theta )\" align=\"absmiddle\"> e&nbsp;<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w=%7Cw%7Ccis%5Calpha\" alt=\"\\dpi{150} \\large w=|w|cis\\alpha\" align=\"absmiddle\">, tal que <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w=%5Csqrt%5Bn%5D%7Bz%7D\" alt=\"\\dpi{150} \\large w=\\sqrt[n]{z}\" align=\"absmiddle\">. Ent&atilde;o, temos que: <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w%5E%7Bn%7D=z\" alt=\"\\dpi{150} \\large w^{n}=z\" align=\"absmiddle\">. Aplicando a primeira f&oacute;rmula de De Moivre para <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w%5E%7Bn%7D\" alt=\"\\dpi{150} \\large w^{n}\" align=\"absmiddle\">:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w%5E%7Bn%7D=%7Cw%7C%5E%7Bn%7Dcis(n%5Calpha&amp;space;)=z=%7Cz%7Ccis%5Ctheta\" alt=\"\\dpi{150} \\large w^{n}=|w|^{n}cis(n\\alpha )=z=|z|cis\\theta\" align=\"absmiddle\">. A igualdade na forma trigonom&eacute;trica implica que:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%7Cw%7C%5E%7Bn%7D=z%5Crightarrow&amp;space;%7Cw%7C=%5Csqrt%5Bn%5D%7Bz%7D\" alt=\"\\dpi{150} \\large |w|^{n}=z\\rightarrow |w|=\\sqrt[n]{z}\" align=\"absmiddle\"> e <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;cos%5Ctheta&amp;space;=&amp;space;cos(n%5Calpha&amp;space;)\" alt=\"\\dpi{150} \\large cos\\theta = cos(n\\alpha )\" align=\"absmiddle\">, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;sen%5Ctheta&amp;space;=sen(n%5Calpha&amp;space;)\" alt=\"\\dpi{150} \\large sen\\theta =sen(n\\alpha )\" align=\"absmiddle\">. Precisamos lembrar que a igualdade entre as fun&ccedil;&otilde;es trigonom&eacute;tricas n&atilde;o implica em igualdade direta entre os &acirc;ngulos, pois as fun&ccedil;&otilde;es s&atilde;o peri&oacute;dicas. Ou seja:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;n%5Calpha&amp;space;=&amp;space;%5Ctheta&amp;space;+&amp;space;2k%5Cpi&amp;space;%5Crightarrow&amp;space;%5Calpha&amp;space;=%5Cfrac%7B%5Ctheta&amp;space;+2k%5Cpi&amp;space;%7D%7Bn%7D\" alt=\"\\dpi{150} \\large n\\alpha = \\theta + 2k\\pi \\rightarrow \\alpha =\\frac{\\theta +2k\\pi }{n}\" align=\"absmiddle\"> . Como os argumentos dos n&uacute;meros complexos s&atilde;o definidos no intervalo <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;%5B0,2%5Cpi&amp;space;)\" alt=\"\\dpi{150} \\large [0,2\\pi )\" align=\"absmiddle\">, s&oacute; faz sentido tomar os valores de k at&eacute; n-1, j&aacute; que para k = n o argumento superaria <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;2%5Cpi\" alt=\"\\dpi{150} \\large 2\\pi\" align=\"absmiddle\">.<\/p><p>Assim:<\/p><p><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/svg.latex?%5Cdpi%7B150%7D&amp;space;%5Clarge&amp;space;w=%5Csqrt%5Bn%5D%7Bz%7D=%5Csqrt%5Bn%5D%7B%7Cz%7C%7Dcis(%5Cfrac%7B0+2k%5Cpi&amp;space;%7D%7Bn%7D)\" alt=\"\\dpi{150} \\large w=\\sqrt[n]{z}=\\sqrt[n]{|z|}cis(\\frac{0+2k\\pi }{n})\" align=\"absmiddle\">, para k = 0,1, &hellip;, n-1.<\/p><p>As f&oacute;rmulas de De Moivre foram nomeadas em homenagem a Abraham De Moivre, matem&aacute;tico franc&ecirc;s com diversos estudos relacionados n&atilde;o s&oacute; aos n&uacute;meros complexos, mas tamb&eacute;m &agrave; trigonometria e &agrave; estat&iacute;stica.<\/p><p>Por conta de conflitos na Fran&ccedil;a referentes a persegui&ccedil;&otilde;es religiosas, De Moivre deixou a Fran&ccedil;a em 1685 e foi para a Inglaterra, onde se tornou membro da Royal Society e continuou a desenvolver seus estudos, principalmente na &aacute;rea de an&aacute;lise de riscos e expectativas.<\/p><p class=\"has-text-align-center has-luminous-vivid-amber-background-color has-background has-medium-font-size\"><a href=\"https:\/\/estrategiavestibulares.com.br\/pacotes\/\" target=\"_blank\">CURSO PARA VESTIBULAR<\/a><\/p><\/p>\n","protected":false},"excerpt":{"rendered":"A F&oacute;rmula de De Moivre &eacute; utilizada para se realizar opera&ccedil;&otilde;es de potencia&ccedil;&atilde;o e de radicia&ccedil;&atilde;o com n&uacute;meros&hellip;\n","protected":false},"author":18,"featured_media":32705,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"wl_entities_gutenberg":"","footnotes":""},"categories":[29],"tags":[],"wl_entity_type":[732],"class_list":{"0":"post-32690","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-matematica","8":"wl_entity_type-article"},"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v25.9 (Yoast SEO v25.9) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>F\u00f3rmula de De Moivre: conceito e como \u00e9 cobrado em prova?<\/title>\n<meta name=\"description\" content=\"F\u00f3rmula de De Moivre: saiba como s\u00e3o calculadas as f\u00f3rmulas de De Moivre e como esse assunto \u00e9 cobrado em vestibulares como o do ITA.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, 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