Prove that if $|z^k |=|z|^k $ then either $k \in \mathbb R$ or $z>0$ Prove that the only circumstances under which $|z^k |=|z|^k $ are (a) When $k $ is a real number or (b) When $z$ is a positive real number and the quantity $(Imk )Log|k|$ is an integral multiple of $2\pi $.
Let $z=|z|e^{i\theta}$ be then $z^k=(|z|e^{i\theta})^k=|z|^k(e^{i\theta})^k $ so $|z^k|=||z|^k(e^{i\theta})^k|=|z|^k|(e^{i\theta})^k|$.  I can conclude that $|(e^{i\theta})^k|=1$? Why?Could you help me with part (b) please? Thank you very much.
 A: If $z$ and $k$ may be complex then we can no longer define $z^k$ in an arithmetic way, but have to resort to the definition
$$z^k:={\rm pv}\bigl(z^k\bigr):=\exp\bigl(k\,{\rm Log}(z)\bigr)\ .$$
Here it is assumed that $k=a+ib$ and that  $z$ does not lie on the negative real axis. Furthermore the principal value of the $\log$ is defined by ${\rm Log}(z):=\log|z|+i{\rm Arg}(z)$, whereby ${\rm Arg}(z)=:\theta\in\>]{-\pi},\pi[\>$ is the (principal value of the) polar angle of $z$. It follows that on the one hand
$$|z|^k=\exp\bigl((a+ib)\log|z|\bigr)\tag{1}$$
and on the other hand
$$\bigl|z^k\bigr|=\exp({\rm Re}\bigl((a+ib)(\log|z|+i\theta)\bigr)\bigr)=\exp\bigl(a\log|z|-b\theta\bigr)\ .\tag{2}$$
The RHSs of $(1)$ and $(2)$ have the same value iff
$$(a+ib)\log|z|=a\log|z|-b\theta+2n\pi i$$
for some $n\in{\mathbb Z}$, and this is the case iff
$$b\theta=0\qquad \wedge\qquad b\log|z|=2n\pi, \ n\in{\mathbb Z}\ .$$
This corresponds to the criteria listed in $(a)$ and $(b)$ of the question, apart from the typo ${\rm Log}(|k|)$ there instead of $\log|z|$.
