Show that $|z^a| = \exp(a \ln |z|) = |z|^a$ when $z\neq{0}$ and $a$ is a real number

Question

Show that if $z\neq{0}$ and $a$ is a real number, then $|z^a| = \exp(a \ln |z|) = |z|^a$, where the principal value of $|z|^a$ is to be taken.

Can someone please help with this problem?

Answer 1

$z^a=\exp\{a\text{log} |z|+i \arg z+2k\pi i\};k\in \Bbb Z$

Since we are taking principal value hence

$z^a=\exp\{a\text{log} |z|+i \text{Arg} z\}\implies |z^a|=\exp\{a\text{log} |z|\}$ since $|e^{ix}|=1 \text {for} $ $x $ real

Hence $|z^a|=\exp\{a\text{log} |z|\}=|z|^a$

Answer 2

Hint: For a complex number you can write it as $z=re^{i\theta}$.