1. Find all complex numbers $z$ for which the equality

$$
z^{2}=\bar{z}
$$

holds.
Solution. Let $z=x+i y(x, y \in \mathbb{R})$. From $z^{2}=\bar{z}$ we get

$$
\begin{aligned}
& x^{2}-y^{2}=x \\
& 2 x y=-y
\end{aligned}
$$

From equation (2) we get $y=0$ or $2 x+1=0$. If $y=0$, then from equation (1) we get $x=0$ or $x=1$. If $2 x+1=0$, i.e., $x=-\frac{1}{2}$, then from equation (1) we get $y=\frac{\sqrt{3}}{2}$ or $y=-\frac{\sqrt{3}}{2}$. Therefore, the required complex numbers are $z_{1}=0, z_{2}=1, z_{3}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, z_{3}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}$.