2B. Find all numbers $n \in \mathbb{N}$ such that the number $z=\left(\frac{3+i}{2-i}\right)^{n}$ is real, where $i$ is the imaginary unit.
Solution. We have

$$
\frac{3+i}{2-i}=\frac{3+i}{2-i} \cdot \frac{2+i}{2+i}=\frac{6+5 i+i^{2}}{4-i^{2}}=1+i
$$

Considering the powers of $1+i$, we conclude that only for $n=4k, k$ being a natural number, $z$ is a real number.