Problem 597

## Solution.

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

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