1A. Determine all natural numbers $n$ for which the number $z=\left(\frac{3+i}{2-i}\right)^{n}$ is real, where $i$ is the imaginary unit.
Solution. The complex number $\frac{3+i}{2-i}$ can be written in the form

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

Thus, $z=(1+i)^{n}$. If $n$ is an even number, i.e., $n=2 k$ for some $k \in \mathbb{N}$, then

$$
z=(1+i)^{n}=(1+i)^{2 k}=\left[(1+i)^{2}\right]^{k}=(2 i)^{k}
$$

If $k$ is an even number, i.e., $k=2 m$ for some $m \in \mathbb{N}$, then

$$
z=(2 i)^{2 m}=\left(4 i^{2}\right)^{m}=4^{m}(-1)^{m}
$$

Thus, if $n=4 m$, i.e., $4 \mid n$, then $z \in \mathbb{R}$.

If $4 \nmid n$, then $n$ has one of the forms $4 m+1, 4 m+2$, or $4 m+3$. In each of these cases, we have

$$
\begin{aligned}
& z=(1+i)^{4 m+1}=4^{m}(-1)^{m}(1+i) \notin \mathbb{R} \\
& z=(1+i)^{4 m+2}=4^{m}(-1)^{m}(1+i)^{2}=4^{m} \cdot(-1)^{m} \cdot 2 i \notin \mathbb{R} \\
& z=(1+i)^{4 m+3}=4^{m}(-1)^{m}(1+i)^{3}=4^{m}(-1)^{m} \cdot 2 i(1+i)=4^{m} \cdot(-1)^{m} \cdot 2(-1+i) \notin \mathbb{R}
\end{aligned}
$$

Thus, $z$ is a real number if and only if $4 \mid n$.