## Task B-2.2.

Determine the complex number $z$ such that

$$
|z+2|=|1-\bar{z}| \quad \text{and} \quad \operatorname{Re}\left(\frac{z}{2+3 i}\right)=\frac{1}{13}
$$
## Solution.

From the first equation $|z+2|=|1-\bar{z}|$ we get $(x+2)^{2}+y^{2}=(1-x)^{2}+y^{2}$, which simplifies to $x=-\frac{1}{2}$

From the second equation, we have

$$
\begin{array}{cc}
\operatorname{Re} \frac{z}{2+3 i}=\operatorname{Re}\left(\frac{x+y i}{2+3 i} \cdot \frac{2-3 i}{2-3 i}\right)=\operatorname{Re}\left(\frac{2 x+3 y}{13}-\frac{3 x-2 y}{13} i\right)=\frac{2 x+3 y}{13}, & 1 \text{ point} \\
\frac{2 x+3 y}{13}=\frac{1}{13} \quad \Rightarrow \quad 2 x+3 y=1 . & 1 \text{ point }
\end{array}
$$

From $2 x+3 y=1$ and $x=-\frac{1}{2}$, it follows that $y=\frac{2}{3}$.

1 point

Therefore, $z=-\frac{1}{2}+\frac{2}{3} i$.

1 point