Exponentiation with complex numbers

Question

If a complex number $z$ is multiplied with itself n times, will the result $z^n$ always be greater in magnitude than $z^{n-1}$ or z ? Is there some formula to find the magnitude of $z^n$ , without finding $z^n$ first?

Answer 1

Note that every complex number $z$ can be written as $re^{i \theta}$ where $r$ is the magnitude of $z$.

Now it follows $z^n=r^n e^{in \theta}$ $$=r^n \left( \cos n \theta + i \sin n \theta \right)$$ $$=r^n \cos n \theta + i r^n \sin n \theta$$

What's the magnitude of $z^n$?

Well, it's $\sqrt{\left(r^n \cos n \theta \right)^2+ \left( r^n \sin n \theta \right)^2}=r^n$

Now ask yourself when $r^n$ is greater than $r^{n-1}$ or, simply $r$.

Answer 2

$|z^n|=|z|^n$.

If $|z|>1$, then the sequence $(|z^n|)$ is strictly increasing,

if $|z|<1$, then the sequence $(|z^n|)$ is strictly decreasing

and if $|z|=1$, then the sequence $(|z^n|)$ is constant.