Let $z \in \mathbb{C}$ and $\bar z$ be its conjugate, $n \in \mathbb{N}$. Solve following equation:
$$\bar z = z^n\qquad (1).$$
My (uncomplete) solution to this:
let $z := a+bi$, where $i$ is imaginary number, and $a,b\in \mathbb{R}$. Then $\bar z = a- bi$.
Using $Moivre's$ formula we get
$$z^n = r^n(\cos n\phi + i\sin n\phi),$$ where $r = |z| = \sqrt{a^2 + b^2}$.
Now we can rewrite $(1)$:
$$a-bi = (a^2+b^2)^{\frac{n}{2}}\cdot(\cos n\phi + i\sin n\phi)$$
Is such approach useful ? I am not if I am steering into right direction, for I'm not sure how solution should look.
Let $z=re^{i\theta}$, then $\overline{z}=re^{-i\theta}$ so the equation is $$re^{-i\theta}=(re^{i\theta})^n$$ $$re^{-i\theta}=r^ne^{in\theta}$$ $$r^{n-1}e^{i(n+1)\theta}=1$$ $$r=1$$ $$\theta=\frac{2\pi k}{n+1}\,\,\, k\in \{0,1,2,...,n\}$$ Edit: As stated in the comments, $z=0$ is also valid.