Problem(3pt). Find all solutions of the given equation:
$$z^5=-i .$$
Let $z=r(cos \theta + isin \theta)$, then $z^5 = r^5[cos(5\theta)+isin(5\theta)]$:
$$-i=1\left[cos\left(\frac{3\pi}{2}+2k\pi\right) +isin\left(\frac{3\pi}{2}+2k\pi\right)\right].$$
Therefore we get
$$r=1^{\frac{1}{5}}=1 \text{ and } \theta=\frac{1}{5}\left(\frac{3\pi}{2}+2k\pi\right) \text{ for } k=0,1,2,3,4$$
The five solutions of the given equation:
\begin{align*}
k&=0: \theta = \frac{3\pi}{10} \Rightarrow z=cos\left(\frac{3\pi}{10}\right) + isin\left(\frac{3\pi}{10}\right)\\
k&=1: \theta = \frac{7\pi}{10} \Rightarrow z=cos\left(\frac{7\pi}{10}\right) + isin\left(\frac{7\pi}{10}\right)\\
k&=2: \theta = \frac{11\pi}{10} \Rightarrow z=cos\left(\frac{11\pi}{10}\right) + isin\left(\frac{11\pi}{10}\right)\\
k&=3: \theta = \frac{3\pi}{2} \Rightarrow z=cos\left(\frac{3\pi}{2}\right) + isin\left(\frac{3\pi}{2}\right)=-i\\
k&=4: \theta = \frac{19\pi}{10} \Rightarrow z=cos\left(\frac{19\pi}{10}\right) + isin\left(\frac{19\pi}{10}\right)
\end{align*}
