Task B-2.7. Let $z_{1}=3\left(\frac{1+i}{1-i}\right)^{2}-2\left(\frac{1-i}{1+i}\right)^{3}$. Determine all complex numbers $z \in \mathbb{C}$ for which

$$
\left|z_{1} \cdot \bar{z}+z \cdot \overline{z_{1}}\right|=4, \quad \operatorname{Im} \frac{z_{1} \cdot z}{1-i}=0
$$
Solution.

$$
z_{1}=3\left(\frac{1+i}{1-i}\right)^{2}-2\left(\frac{1-i}{1+i}\right)^{3}=3 \cdot \frac{1+2 i-1}{1-2 i-1}-2 \cdot \frac{1-3 i-3+i}{1+3 i-3-i}=-3-2 \cdot \frac{-2-2 i}{-2+2 i}=-3-2 i
$$

Let $z=x+y i$. We need to solve the system of equations

$$
\begin{array}{r}
|(-3-2 i) \cdot(x-y i)+(x+y i) \cdot(-3+2 i)|=4 \\
\operatorname{Im} \frac{(-3-2 i) \cdot(x+y i)}{1-i}=0
\end{array}
$$

After performing the operations, the system reduces to the form:

$$
|-6 x-4 y|=4, \quad \frac{5 x+y}{2}=0
$$

From the last equation, we get $y=-5 x$ so $|14 x|=4$.

There are two possibilities:

$$
x=\frac{2}{7}, y=-\frac{10}{7} \quad \text { or } \quad x=-\frac{2}{7}, y=\frac{10}{7}
$$

Then $z=\frac{2}{7}-\frac{10}{7} i$ or $z=-\frac{2}{7}+\frac{10}{7} i$.