Find all complex numbers

Question

$\forall{n}\in\mathbb{N}$ find all complex numbers $z$ satisfying the equation $z^n=-\bar{z}$. I found $0$ and $1$ and for other? How can I prove it?

Answer 1

$\displaystyle z^n = -\overline z$.

Let $\displaystyle z = re^{i\theta}$, giving $\displaystyle r^ne^{in\theta} = -re^{-i\theta}$

Since the modulus of the two sides has to be equal, $\displaystyle r^n = r$, giving the solutions $\displaystyle r=0$ and $\displaystyle r=1$. $\displaystyle r=0 \implies z=0$ is a single solution, while $\displaystyle r = 1 \implies z = e^{i\theta}$ is a set of solutions we need to find.

We have $\displaystyle e^{in\theta} = -e^{-i\theta}$

$\displaystyle e^{i\theta(n+1)} = -1$

$\displaystyle e^{i\theta(n+1)} = e^{(2k+1)i\pi}, \forall k \in \mathbb{Z}$

Hence $\displaystyle \theta = \frac{2k+1}{n+1}\pi, \forall k \in \mathbb{Z}$

giving $\displaystyle z = e^{i(\frac{2k+1}{n+1})\pi} = \cos (\frac{2k+1}{n+1})\pi + i\sin (\frac{2k+1}{n+1})\pi , \forall k \in \mathbb{Z}$

To eliminate duplicate solutions, consider the argument modulo $\displaystyle 2\pi$.

Answer 2

If $z^n=-\overline z$, then $z^{n+1}=-\overline zz=-|z|^2$. So, $|z|^{n+1}=|z|^2$ and therefore $|z|=0(\iff z=0)$ or $|z|=1$, unless $n=1$ (but the equation $z=-\overline z$ is easy to solve: its solutions are the purely imaginary numbers). So, unless $z=0$, you know that $z=e^{i\theta}$ for some $\theta\in\Bbb R$. And then your equation becomes $e^{in\theta}=e^{-i\theta}$. Can you take it from here?