Task B-2.2. Determine all complex numbers $z$ such that $z^{3}=\bar{z}$.
Solution. Let $z=x+y i, x, y \in \mathbb{R}$. Then we have

$$
\begin{aligned}
(x+y i)^{3} & =x-y i \\
x^{3}+3 x^{2} y i-3 x y^{2}-y^{3} i & =x-y i
\end{aligned}
$$

For the equality to hold, the real and imaginary parts on both sides must be equal.

$$
x^{3}-3 x y^{2}=x \quad \text { and } \quad 3 x^{2} y-y^{3}=-y .
$$

We obtain the system of equations

$$
\begin{aligned}
& x^{3}-3 x y^{2}-x=0 \\
& 3 x^{2} y-y^{3}+y=0
\end{aligned}
$$

From (1) it follows that

$$
\begin{aligned}
& x\left(x^{2}-3 y^{2}-1\right)=0 \\
& x=0 \quad \text { or } \quad x^{2}-3 y^{2}-1=0
\end{aligned}
$$

If $x=0$, then from (2) we have

$$
-y^{3}+y=0
$$

or $-y\left(y^{2}-1\right)=0$ so $y=0$ or $y=1$ or $y=-1$.

If $x^{2}-3 y^{2}-1=0$, i.e., $x^{2}=3 y^{2}+1$, then from (2) it follows that

$$
3\left(3 y^{2}+1\right) y-y^{3}=-y
$$

or $8 y^{3}+4 y=0$, which is possible only for $y=0$.

Then $x^{2}-1=0$ so $x=1$ or $x=-1$.

Thus, the complex numbers $z$ for which the given equality holds are

$$
z_{1}=0, \quad z_{2}=i, \quad z_{3}=-i, \quad z_{4}=1, \quad z_{5}=-1
$$