Complex numer equation

Question

Let $n\in\mathbb{N}$. Determine all complex numbers $z\in\mathbb{C}$ such that $z^{n-1}$ = $\bar{z}$ .

I'm not sure if I'm doing this question right, but would the solutions be $+ 1,-1$ or $0$?

Answer 1

Well, partly, yes, these are solutions (unless $n=1$ and $z=0$) but there might be other solutions as well.

Use polar coordinates, i.e. assume that $z$ has distance $r$ from $0$ (i.e. $|z|=r$) and it encloses angle $\alpha$ with the right wing of the real axis. Then $z^{n-1}$ will have length $r^{n-1}$ and angle $(n-1)\alpha$, while $\bar z$ has length $r$ and angle $-\alpha$.