Task B-4.2. Solve the equation $z^{5} i+(1-i) \cdot z^{2}=0$ in the set $\mathbb{C}$.
Solution.

$$
\begin{aligned}
& z^{5} i+(1-i) \cdot z^{2}=0 \\
& z^{2}\left(z^{3} i+(1-i)\right)=0 \\
& z^{2}=0 \Leftrightarrow z=0 \text { (double solution) } \\
& z^{3} i+(1-i)=0 \\
& z^{3}=\frac{-1+i}{i}=1+i \\
& z^{3}=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \\
& z_{k}=\sqrt[6]{2}\left(\cos \frac{\frac{\pi}{4}+2 k \pi}{3}+i \sin \frac{\frac{\pi}{4}+2 k \pi}{3}\right), \quad k=0,1,2
\end{aligned}
$$

The solutions to the equation are:

$$
\begin{aligned}
& z_{0}=\sqrt[6]{2}\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right) \\
& z_{1}=\sqrt[6]{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right) \\
& z_{2}=\sqrt[6]{2}\left(\cos \frac{17 \pi}{12}+i \sin \frac{17 \pi}{12}\right)
\end{aligned}
$$

and the double solution $z=0$.