How big is $|z^w|$ compared to $|z|$?

Question

I've found a nice way to write complex exponentiation, and when we look at that formula, $|z^w|$ and $Arg(z^w)$ become much more apparent. So using the formula, with what kind of proportionality can we describe $|z^w|$ compared to $|z|$? $$z^w = \frac{|z|^{Re(w)}}{e^{Arg(z)Im(w)}}e^{(Arg(z)Re(w)+ln(|z|)Im(w))i}$$If $Re(w)$ is $0$, one can see that $Arg(z^w)$ is logarithmically proportional to $|z|$, but I do not know how to describe the proportionality of $Arg(z)$ compared to $|z^w|$, but I suspect one could say that $|z^w|$ is negatively exponentially proportional to $Arg(z)$ or something like that.

Answer 1

Let $z = re^{j \theta}$, $w = a+jb$, and $j=\sqrt{-1}$.

By definition, $|z|=r$.

$z^w = z^{a+jb}$

= $z^a z^{jb}$

= $r^a e^{ja\theta} r^{jb} e^{-\theta b}$

We know that $|e^{jab}|=1$ and also $|r^{jb}| =1$ because you can rewrite $r^{jb}=e^{jb'}$.

This leaves us with $|z^w| = r^a e^{-\theta b}$. In other words, $|z^w| = |z|^{\Re{w}} e^{-\Im{w} \cdot Arg{z}}$, where $\Re$ and $\Im$ denote the real and imaginary parts, respectively.

Seems hard to simplify beyond this for the most general case.

Answer 2

$$\left|z^w\right|=\left|e^{w\log z}\right|=e^{\Re(w\log z)}=e^{\Re(w)\log|z|-\Im(w)\arg(z)}=|z|^{\Re(w)}e^{-\Im(w)\arg(z)}.$$

But notice that the argument is not uniquely defined, you need to choose a branch.

Obviously, for real $w$,

$$|z^w|=|z|^w.$$