Problem 882

## First Solution.

Let $z=r(\cos \varphi+i \sin \varphi), r=|z|, \varphi \in\left\langle\frac{3 \pi}{2}, 2 \pi\right\rangle$.

Since $|z+i z|=|z(1+i)|=|z| \cdot|1+i|=r \cdot \sqrt{2}=2$, we conclude that $r=\sqrt{2}. \quad 1$ point

Furthermore, $z^{4}=r^{4}(\cos 4 \varphi+i \sin 4 \varphi)$, and thus $\operatorname{Re}\left(z^{4}\right)=r^{4} \cos 4 \varphi=-2$.

By substituting $r=\sqrt{2}$ into $r^{4} \cos 4 \varphi=-2$, we get $\cos 4 \varphi=-\frac{1}{2}$, from which we conclude that $4 \varphi= \pm \frac{2 \pi}{3}+k \cdot 2 \pi, k \in \mathbf{Z}$, or that $\varphi= \pm \frac{\pi}{6}+k \cdot \frac{\pi}{2}, k \in \mathbf{Z}$.

By substituting $k \in \mathbf{Z}$ and considering the fact that $\varphi \in\left\langle\frac{3 \pi}{2}, 2 \pi\right\rangle$, we conclude that $\varphi=\frac{5 \pi}{3}$ or $\varphi=\frac{11 \pi}{6}$.

Therefore, the solutions are:

$$\begin{gathered} z_{1}=\sqrt{2}\left(\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right)=\sqrt{2}\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)=\frac{\sqrt{2}}{2}-i \frac{\sqrt{6}}{2} \\ z_{2}=\sqrt{2}\left(\cos \frac{11 \pi}{6}+i \sin \frac{11 \pi}{6}\right)=\sqrt{2}\left(\frac{\sqrt{3}}{2}-i \frac{1}{2}\right)=\frac{\sqrt{6}}{2}-i \frac{\sqrt{2}}{2} \end{gathered}$$

Note: The student can leave the solution in trigonometric or standard form.