Exercício Resolvido de Números Complexos
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d) \( \sqrt[{3}]{-i} \)
As raízes de um número complexo são dadas por
\[ \bbox[#99CCFF,10px]
{z=\sqrt[{n}]{r}\left(\cos \frac{\theta +2k\pi}{n}+i\operatorname{sen}\frac{\theta +2k\pi }{n}\right)}
\]
\[
\begin{gather}
x+iy=0-i\\[5pt]
r=\sqrt{0^{2}+(-1)^{2}\;}=\sqrt{1\;}=1
\end{gather}
\]
\[
\begin{gather}
\theta=\operatorname{arctg}\left(\frac{-1}{0}\right)=\operatorname{arctg}(-\infty)=\frac{3\pi }{2}
\end{gather}
\]
Para a raiz cúbica, n=3 e k=0, 1, 2.
Para k=0:
\[
\begin{gathered}
z_{1}=\sqrt[{3}]{1}\left(\cos \frac{\frac{3\pi}{2}+2.0\pi }{3}+i\operatorname{sen}\frac{\frac{3\pi }{2}+2.0\pi}{3}\right)\\
z_{1}=1\left[\cos \left(\frac{\pi}{2}\right)+i\operatorname{sen}\left(\frac{\pi}{2}\right)\right]\\
z_{1}=1\left(0+i\right)
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{1}=i}
\]
Para k=1:
\[
\begin{gathered}
z_{2}=\sqrt[{3}]{1}\left(\cos \frac{\frac{3\pi}{2}+2.1\pi }{3}+i\operatorname{sen}\frac{\frac{3\pi }{2}+2.1\pi}{3}\right)\\
z_{2}=1\left(\cos \frac{\frac{3\pi +4\pi}{2}}{3}+i\operatorname{sen}\frac{\frac{3\pi +4\pi}{2}}{3}\right)\\
z_{2}=1\left(\cos \frac{7\pi}{6}+i\operatorname{sen}\frac{7\pi}{6}\right)\\
z_{2}=1\left(-{\frac{\sqrt{3}}{2}}-i\frac{1}{2}\right)
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{2}=-{\frac{\sqrt{3}}{2}}-i\frac{1}{2}}
\]
Para k=2:
\[
\begin{gathered}
z_{3}=\sqrt[{3}]{1}\left(\cos \frac{\frac{3\pi}{2}+2.2\pi }{3}+i\operatorname{sen}\frac{\frac{3\pi }{2}+2.2\pi}{3}\right)\\
z_{3}=1\left(\cos \frac{\frac{3\pi +8\pi}{2}}{3}+i\operatorname{sen}\frac{\frac{3\pi +8\pi}{2}}{3}\right)\\
z_{3}=1\left(\cos \frac{11\pi}{6}+i\operatorname{sen}\frac{11\pi}{6}\right)\\
z_{3}=1\left(\frac{\sqrt{3}}{2}-i\frac{1}{2}\right)
\end{gathered}
\]
\[ \bbox[#FFCCCC,10px]
{z_{3}=\frac{\sqrt{3}}{2}-i\frac{1}{2}}
\]
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