Prove that $(z^a)^b = z^{ab}e^{2kb\pi i}$

Question

I'm trying to prove that $$ (z^a)^b = z^{ab}e^{2kb\pi i}, $$ where $a,b,z\in\mathbb{C}$ and $k\in\mathbb{Z}$.

So far, I have that \begin{align*} (z^a)^b & = (\exp(a\log z))^b \\ & = \exp(b\log(\exp(a\log z))), \end{align*} and I know that the complex logarithm can be defined such that $\log z=\log|z|+i\theta+2k\pi i$, where $\theta\in(-\pi,\pi]$ is the principal argument of $z$.

I feel like I've got all the components I need, but I'm not quite sure how to piece it together to get the desired result.

Answer 1

Instead use $(z^a)^b=(e^{a\ln z+2k\pi i})^b=e^{ab\ln z+2kb\pi i}=e^{ab\ln z}e^{2kb\pi i}=z^{ab}e^{2kb\pi i}$.