13. Solve the equation $z^{n}=(\bar{z})^{m}\left(m, n \in \mathbf{N}^{*}\right)$.
13. When $m=n$, it is obvious that $z \in \mathbf{R}$ is a solution to the equation; when $z$ is an imaginary number, let $z=r(\cos \theta+i \sin \theta), r>0$, then we have $(\cos \theta+\mathrm{i} \sin \theta)^{2 n}=1$. Therefore, $\cos \theta+\mathrm{i} \sin \theta= \pm\left(\cos \frac{k \pi}{n}+\mathrm{i} \sin \frac{k \pi}{n}\right), k=1,2, \cdots, n-1, n \geqslant 2$. At this time, $z=$ $\pm r\left(\cos \frac{k \pi}{n}+\mathrm{i} \sin \frac{k \pi}{n}\right), k=1,2, \cdots, n-1, n \geqslant 2$.

When $m \neq n$, taking the modulus on both sides of the equation, we get $|z|^{n}=|\bar{z}|^{m}$. Without loss of generality, assume $n>m$. Clearly, $z=0$ is a solution to the equation; when $z \neq 0$, we have $|z|^{n-m}=1$. Therefore, $|z|=1$, and thus $z \bar{z}=1$, so $z^{m-n}=1$. At this time, $z=\cos \frac{2 k \pi}{m+n}+i \sin \frac{2 k \pi}{m+n}, k=0,1, \cdots, m+n-1$