## Task B-2.2.

Let $z=x+2i$ and $u=3+yi$ where $(x, y \in \mathbb{R})$ be complex numbers. Determine all ordered pairs of real numbers $(x, y)$ for which the real part of the number $\frac{z+u}{z-u}$ is zero, and the point $s$ with coordinates $(x, y)$ is 3 units away from the origin.
## Solution.

Determine the real part of the number $\frac{z+u}{z-u}$.

$$
\begin{aligned}
& \frac{z+u}{z-u}=\frac{x+2 i+3+y i}{x+2 i-3-y i}=\frac{(x+3)+(2+y) i}{(x-3)+(2-y) i} \\
&=\frac{((x+3)+(2+y) i)((x-3)-(2-y) i)}{(x-3)^{2}+(2-y)^{2}} \\
& \operatorname{Re}\left(\frac{z+u}{z-u}\right)=\frac{(x+3)(x-3)+(2+y)(2-y)}{(x-3)^{2}+(2-y)^{2}}=\frac{x^{2}-y^{2}-5}{(x-3)^{2}+(2-y)^{2}}
\end{aligned}
$$

This fraction is equal to zero if $x^{2}-y^{2}-5=0$ with $x \neq 3, y \neq 2$.

The point with coordinates $(x, y)$ is 3 units away from the origin if $x^{2}+y^{2}=9$.

The solution to the problem is all ordered pairs $(x, y)$ that satisfy the system

$$
\begin{aligned}
& x^{2}-y^{2}=5 \\
& x^{2}+y^{2}=9
\end{aligned}
$$

If we add the equations of this system, we get $x^{2}=7$ so $x= \pm \sqrt{7}$. Then $y^{2}=2$ or $y= \pm \sqrt{2}$.

Finally, the solutions are the ordered pairs

$$
(-\sqrt{7},-\sqrt{2}),(-\sqrt{7}, \sqrt{2}),(\sqrt{7},-\sqrt{2}),(\sqrt{7}, \sqrt{2})
$$