## Task B-4.2.

Let $z=x+2i$ and $w=3+yi$ be complex numbers, where $x, y \in \mathbb{R}$. Determine the smallest positive real number $x$ for which the fraction $\frac{z+w}{z-w}$ is an imaginary number.
## Solution.

The complex number $\frac{z+w}{z-w}$ is imaginary if and only if its real part is zero. 1 point Therefore, let's determine $\operatorname{Re}\left(\frac{z+w}{z-w}\right)$.

$$
\begin{aligned}
\frac{z+w}{z-w} & =\frac{x+2 i+3+y i}{x+2 i-3-y i}=\frac{x+3+(2+y) i}{x-3+(2-y) i} \cdot \frac{x-3-(2-y) i}{x-3-(2-y) i} & & 1 \text { point } \\
& =\frac{\left(x^{2}-3^{2}\right)+\left(2^{2}-y^{2}\right)+i((x-3)(2+y)-(x+3)(2-y))}{(x-3)^{2}+(2-y)^{2}} & & 1 \text { point }
\end{aligned}
$$

Then we have

$$
\operatorname{Re}\left(\frac{z+w}{z-w}\right)=\frac{x^{2}-y^{2}-5}{(x-3)^{2}+(2-y)^{2}}
$$

This expression is equal to zero if and only if the numerator is zero, i.e., $x^{2}-y^{2}-5=0$.

From this, we get $x^{2}=y^{2}+5$, and the smallest positive $x$ is obtained for $y=0$ and is $x=\sqrt{y^{2}+5}=\sqrt{5}$.