Exercício Resolvido de Números Complexos
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e) \( \left(-1+i\sqrt{3}\right)^{1/4} \)
As raízes de um número complexo são dadas por
\[ \bbox[#99CCFF,10px]
{z=\sqrt[{n}]{r}\left(\cos \frac{\theta +2k\pi}{n}+i\operatorname{sen}\frac{\theta +2k\pi }{n}\right)}
\]
\[
\begin{gather}
x+iy=-1+i\sqrt{3}\\[5pt]
r=\sqrt{(-1)^{2}+\left(\sqrt{3}\right)^{2}\;}=\sqrt{1+3\;}=\sqrt{4\;}=2
\end{gather}
\]
\[
\begin{gather}
\theta=\operatorname{arctg}\left(\frac{\sqrt{3}}{-1}\right)=\operatorname{arctg}\left(-\sqrt{3}\right)=\frac{2\pi}{3}
\end{gather}
\]
Para a raiz quarta, n=4 e k=0, 1, 2, 3.
Para k=0:
\[
\begin{gathered}
z_{1}=\sqrt[{4}]{2}\left(\cos \frac{\frac{2\pi}{3}+2.0\pi }{4}+i\operatorname{sen}\frac{\frac{2\pi }{3}+2.0\pi}{4}\right)\\
z_{1}=\sqrt[{4}]{2}\left[\cos \left(\frac{2\pi}{12}\right)+i\operatorname{sen}\left(\frac{2\pi}{12}\right)\right]\\
z_{1}=\sqrt[{4}]{2}\left[\cos \left(\frac{\pi}{6}\right)+i\operatorname{sen}\left(\frac{\pi}{6}\right)\right]\\
z_{1}=\sqrt[{4}]{2}\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)
\end{gathered}
\]
multiplicando o numerador e o denominador por
\( \sqrt[{4}]{2^{3}} \)
\[
\begin{gathered}
z_{1}=\sqrt[{4}]{2}\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right).\frac{\sqrt[{4}]{2^{3}}}{\sqrt[{4}]{2^{3}}}\\
z_{1}=\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right).\frac{2^{1/4}.2^{3/4}}{\sqrt[{4}]{2^{3}}}\\
z_{1}=\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right).\frac{2^{4/4}}{\sqrt[{4}]{2^{3}}}\\
z_{1}=\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right).\frac{2}{\sqrt[{4}]{8}}
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{1}=\frac{\sqrt{3}}{\sqrt[{4}]{8}}+i\frac{1}{\sqrt[{4}]{8}}}
\]
Para k=1:
\[
\begin{gathered}
z_{2}=\sqrt[{4}]{2}\left(\cos \frac{\frac{2\pi}{3}+2.1\pi }{4}+i\operatorname{sen}\frac{\frac{2\pi }{3}+2.1\pi}{4}\right)\\
z_{2}=\sqrt[{4}]{2}\left[\cos \left(\frac{8\pi}{12}\right)+i\operatorname{sen}\left(\frac{8\pi}{12}\right)\right]\\
z_{2}=\sqrt[{4}]{2}\left[\cos \left(\frac{2\pi}{3}\right)+i\operatorname{sen}\left(\frac{2\pi}{3}\right)\right]\\
z_{2}=\sqrt[{4}]{2}\left(-{\frac{1}{2}}+i\frac{\sqrt{3}}{2}\right)
\end{gathered}
\]
multiplicando o numerador e o denominador por
\( \sqrt[{4}]{2^{3}} \)
\[
\begin{gathered}
z_{2}=\sqrt[{4}]{2}\left(-{\frac{1}{2}}+i\frac{\sqrt{3}}{2}\right).\frac{\sqrt[{4}]{2^{3}}}{\sqrt[{4}]{2^{3}}}\\
z_{2}=\left(-{\frac{1}{2}}+i\frac{\sqrt{3}}{2}\right).\frac{2}{\sqrt[{4}]{8}}
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{2}=-{\frac{1}{\sqrt[{4}]{8}}}+i\frac{\sqrt{3}}{\sqrt[{4}]{8}}}
\]
Para k=2:
\[
\begin{gathered}
z_{3}=\sqrt[{4}]{2}\left(\cos \frac{\frac{2\pi}{3}+2.2\pi }{4}+i\operatorname{sen}\frac{\frac{2\pi }{3}+2.2\pi}{4}\right)\\
z_{3}=\sqrt[{4}]{2}\left[\cos \left(\frac{14\pi}{12}\right)+i\operatorname{sen}\left(\frac{14\pi}{12}\right)\right]\\
z_{3}=\sqrt[{4}]{2}\left[\cos \left(\frac{7\pi}{6}\right)+i\operatorname{sen}\left(\frac{7\pi}{6}\right)\right]\\
z_{3}=\sqrt[{4}]{2}\left(-{\frac{\sqrt{3}}{2}}-i\frac{1}{2}\right)
\end{gathered}
\]
multiplicando o numerador e o denominador por
\( \sqrt[{4}]{2^{3}} \)
\[
\begin{gathered}
z_{3}=\sqrt[{4}]{2}\left(-{\frac{\sqrt{3}}{2}-i\frac{1}{2}}\right).\frac{\sqrt[{4}]{2^{3}}}{\sqrt[{4}]{2^{3}}}\\
z_{3}=\left(-{\frac{\sqrt{3}}{2}-i\frac{1}{2}}\right).\frac{2}{\sqrt[{4}]{8}}
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{3}=-{\frac{\sqrt{3}}{\sqrt[{4}]{8}}}-i\frac{1}{\sqrt[{4}]{8}}}
\]
Para k=3:
\[
\begin{gathered}
z_{4}=\sqrt[{4}]{2}\left(\cos \frac{\frac{2\pi}{3}+2.3\pi }{4}+i\operatorname{sen}\frac{\frac{2\pi }{3}+2.3\pi}{4}\right)\\
z_{4}=\sqrt[{4}]{2}\left[\cos \left(\frac{20\pi}{12}\right)+i\operatorname{sen}\left(\frac{20\pi}{12}\right)\right]\\
z_{4}=\sqrt[{4}]{2}\left[\cos \left(\frac{10\pi}{6}\right)+i\operatorname{sen}\left(\frac{10\pi}{6}\right)\right]\\
z_{4}=\sqrt[{4}]{2}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)
\end{gathered}
\]
multiplicando o numerador e o denominador por
\( \sqrt[{4}]{2^{3}} \)
\[
\begin{gathered}
z_{4}=\sqrt[{4}]{2}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right).\frac{\sqrt[{4}]{2^{3}}}{\sqrt[{4}]{2^{3}}}\\
z_{4}=\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right).\frac{2}{\sqrt[{4}]{8}}
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{4}=\frac{1}{\sqrt[{4}]{8}}-i\frac{\sqrt{3}}{\sqrt[{4}]{8}}}
\]
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Fisicaexe - Exercícios Resolvidos de Física de Elcio Brandani Mondadori está licenciado com uma Licença Creative Commons - Atribuição-NãoComercial-Compartilha Igual 4.0 Internacional .