9. (16 points) Let $n \geqslant 3$ be a given positive odd number, and let $z=\cos \frac{2 \pi}{n}+\mathrm{i} \sin \frac{2 \pi}{n}$ be fixed. Consider the sets:
$$
\begin{array}{l}
A=\left\{z, z^{2}, \cdots, z^{n-1}\right\}, \\
B=\left\{1+z, 1+z+z^{2}, \cdots, 1+z+\cdots+z^{n-1}\right\} .
\end{array}
$$

Find $A \cap B$.
9. Let $s \in A \cap B, s \neq 1$. Then
$$
\begin{array}{l}
s=1+z+z^{2}+\cdots+z^{k} \\
=\frac{1-z^{k+1}}{1-z}(k=1,2, \cdots, n-1) .
\end{array}
$$

Since $s \in A$, we know $|s|=1$.
Thus $\left|1-z^{k+1}\right|=|1-z|$
$$
\begin{array}{l}
\Rightarrow \sin \frac{(k+1) \pi}{n}=\sin \frac{\pi}{n} \\
\Rightarrow \frac{(k+1) \pi}{n}=\pi-\frac{\pi}{n} \\
\Rightarrow k=n-2 \\
\Rightarrow s=\frac{1-\frac{1}{z}}{1-z}=-\frac{1}{z} \\
\Rightarrow s^{n}=-1 .
\end{array}
$$

This contradicts $s \in A$.
Therefore, $A \cap B=\varnothing$.