## Task A-4.1.

Determine all complex numbers $z$ for which $z^{2}=(1+i) \cdot \bar{z}$.
## First Solution.

Let $z=x+i y$, where $x$ and $y$ are real numbers. Then we have

$$
z^{2}=x^{2}-y^{2}+2 i x y, \quad \text { and } \quad(1+i) \bar{z}=(1+i)(x-i y)=x+y+i(x-y). \quad 1 \text { point }
$$

Substituting into the initial equation and comparing the real and imaginary parts of the complex number, we get the system

$$
\begin{array}{r}
x^{2}-y^{2}=x+y \\
2 x y=x-y
\end{array}
$$

From the first equation, we have $(x-y)(x+y)=x+y$, which simplifies to

$$
(x-y-1)(x+y)=0
$$

In the case where $x+y=0$, we can express $y=-x$ and substitute it into the second equation of the system.

We get $-2 x^{2}=2 x$, from which we obtain the solutions $x=0,-1, \text{ i.e. } z=0,-1+i$.

In the case where $x-y-1=0$, we express $y=x-1$ and substitute it back into the second equation.

We get the equation $2 x(x-1)=1$ whose solutions are $x=\frac{1+\sqrt{3}}{2}, \frac{1-\sqrt{3}}{2}$, i.e.,

$$
z=\frac{1+\sqrt{3}}{2}+i \frac{-1+\sqrt{3}}{2}, \quad \frac{1-\sqrt{3}}{2}+i \frac{-1-\sqrt{3}}{2}
$$

Therefore, the solutions to the initial equation are

$$
z=0, \quad-1+i, \quad \frac{1+\sqrt{3}}{2}+i \frac{-1+\sqrt{3}}{2}, \quad \frac{1-\sqrt{3}}{2}+i \frac{-1-\sqrt{3}}{2}
$$