Task B-4.1. Determine all complex numbers $z$ for which

$$
z^{4}+2 \cdot(1-i) \cdot z^{2}-2 i=0
$$
## Solution.

We introduce the substitution $z^{2}=t$. Then the given equation transforms into

$$
\begin{gathered}
t^{2}+2 \cdot(1-i) \cdot t-2 i=0 \\
t_{1,2}=\frac{-2(1-i) \pm \sqrt{4 \cdot(-2 i)+8 i}}{2}=\frac{-2+2 i}{2} \\
z^{2}=-1+i \\
z=\sqrt{-1+i}, \quad-1+i=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right) \\
z_{1}=\sqrt[4]{2}\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right), \quad z_{2}=\sqrt[4]{2}\left(\cos \frac{11 \pi}{8}+i \sin \frac{11 \pi}{8}\right)
\end{gathered}
$$