Determine all complex numbers z in equation:

Question

Let $n\in\mathbb{N}$. Determine all complex numbers $z\in\mathbb{C}$ such that $z^{n-1}$ = $\bar{z}$
How would I begin this?
Would I begin by saying $z=a+ib$ and expand and stuff?

Answer 1

Certainly $z = 0$ is a solution, so suppose $z \ne 0$. Multiplying both sides by $z$, it is sufficient to study

$$z^n = z \overline{z} = |z|^2$$

Thus $z^n$ needs to be real, and taking absolute values shows that

$$|z|^2 = |z^n| = |z|^n \implies |z|^{n - 2} = 1$$

So unless $n = 2$, it's necessary that $z$ have modulus $1$, in which case we can write

$$z = e^{it}$$

for some $0 \le t < 2\pi$. Now it's just a matter of computation.

If $n = 2$, though, the original equation reads $z = \overline{z}$, which is equivalent to requiring that $z$ is real.

Answer 2

Let $\implies z=re^{i\phi}\implies\bar z=re^{-i\phi}$

So using de Moivre's Theorem we have $ \displaystyle r^ne^{i n\phi}=re^{-i\phi}$

Clearly $r=0\iff z=0$ is a solution,

else we have $\displaystyle r^{n-1}e^{i n\phi}=e^{-i\phi}$

Taking modulus, $\displaystyle r^{n-1}=1$

either $r=1\implies e^{i n\phi}=e^{-i\phi}\iff e^{i(n+1)\phi}=1=e^{2m\pi i}$ where $m$ is any integer

$\displaystyle\implies (n+1)\phi=2m\pi$

$\displaystyle\implies z=1\cdot e^{\left(i\dfrac{2m\pi}{n+1}\right)}$

or $n-1=0\iff n=1\implies e^{i\phi}=e^{-i\phi}\iff e^{2i\phi}=1=e^{2s\pi i}$ where $s$ is any integer

$\displaystyle\implies\phi=s\pi$

Here $r$ can assume any non-zero real value

$\displaystyle\implies z=r\cdot e^{i s\pi}=r (e^{i\pi})^s=r(-1)^s$