## Task A-4.1.

Determine all complex numbers $z$ that satisfy the equations

$$
|z+1|=1 \quad \text{and} \quad\left|z^{2}+1\right|=1
$$
## First Solution.

Let $z=a+bi$. We have $z^{2}=(a+bi)(a+bi)=\left(a^{2}-b^{2}\right)+2abi$.

From the (squared) conditions of the problem, we get

$$
\begin{array}{r}
(a+1)^{2}+b^{2}=1 \\
\left(a^{2}-b^{2}+1\right)^{2}+4a^{2}b^{2}=1
\end{array}
$$

From the first equation, we get $b^{2}=-a^{2}-2a$.

We substitute this into the second equation:

$$
\begin{aligned}
\left(2a^{2}+2a+1\right)^{2}-4a^{2}\left(a^{2}+2a\right) & =1 \\
4a^{4}+4a^{2}+1+8a^{3}+4a^{2}+4a-4a^{4}-8a^{3} & =1 \\
8a^{2}+4a & =0
\end{aligned}
$$

The solutions to this equation are $a=0$ and $a=-\frac{1}{2}$. For these, we get the corresponding values for the imaginary part of the number $z: b=0$ and $b= \pm \frac{\sqrt{3}}{2}$.

1 point

Thus, the solutions to the system are $z_{1}=0$ and $z_{2,3}=\frac{-1 \pm \sqrt{3}i}{2}$.

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