Determining all complex Z in the equation

Question

Let $n \in \mathbb N$. Determine all complex numbers $z \in \mathbb C $ such that $ |z| ^{n-2} = 1.$
How would i begin this question, thanks!

Answer 1

Let $r=|z|$ then notice that $r$ is a nonnegative real number.

$r^{n-2}=1$ if $n=2$ then it is true for all $r>0\implies $ it is true for all $z\neq 0$

if $n\ne2$ then $r=1\implies$ it is true for all $z=xi+y$ such that $x^2+y^2=1$

Answer 2

If $n = 2$, then $\vert z \vert = 1$ for all $z \ne 0$, and perhaps also for $z = 0$, depending on how one looks at $0^0$. In any event, the case $n =2$ stands by itself, especially in light of the fact that $\vert z \vert^{n -2} = 1$ implies $\vert z \vert = 1$ for $n \ne 2$; then any unimodular $z$ is a solution, and only such $z$ are solutions. Thus $z = e^{i \theta}$ for any real $\theta$; of course, since $e^{i(\theta + 2\pi)} = e^{i \theta}$, we only need consider $\theta \in [0, 2\pi)$; such $\theta$ yield all solutions to $\vert z \vert^{n - 2} = 1$ when $n \ne 2$.

Hope this helps. Cheerio,

and as always, Fiat Lux!!!**