## Task A-2.2.

Determine all complex numbers $z$ for which

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

If we substitute $z=x+y i, x, y \in \mathbb{R}$, the conditions become $\left|x^{2}+2 x y i-y^{2}-i\right|=1$ and $|x+y i|=\sqrt{2}$, which is equivalent to

$$
\left(x^{2}-y^{2}\right)^{2}+(2 x y-1)^{2}=1, \quad x^{2}+y^{2}=2
$$

Since it holds that

$$
\left(x^{2}-y^{2}\right)^{2}+(2 x y-1)^{2}=\left(x^{2}+y^{2}\right)^{2}-4 x y+1
$$

from the given conditions, it follows that $1=2^{2}-4 x y+1$, i.e., $x y=1$.

It remains to solve the system

$$
\begin{gathered}
x y=1 \\
x^{2}+y^{2}=2 .
\end{gathered}
$$

Notice that

$$
(x-y)^{2}=x^{2}+y^{2}-2 x y=2-2 \cdot 1=0
$$

Therefore, $x=y$ and substituting into the first equation of the system gives $x^{2}=1$.

We have two solutions to the system: $x=y=1$ and $x=y=-1$, i.e., the desired numbers are $z=1+i$ and $z=-1-i$.