Please let me know if this is a duplicate. I couldn't find any related questions so far.
In my class, we often find the branch cut of $f(z) = z^k$ by equating the function to $f(z) = e^{Log(z^k)} = e^{kLog(z)}$.
An example can be seen here (prof's note):
However, we cannot say ${Log(z^k)} = {kLog(z)}$ (for example, take $Log_\pi ((-1)^2)$).
I've been trying to prove the equation $e^{Log_*(z^k)} = e^{kLog_*(z)} $ or to find the condition under which it holds.
Following is my unsuccessful proof so far:
Let $z = re^{i\theta}, r\in \mathbb{R}/\mathbb{R^-}, \theta \in \mathbb{R}$.
Hence, $z^k = r^ke^{ik\theta}$
$Log_\pi (z^k) = \ln|r^ke^{ik\theta}| + iArg_\pi(r^ke^{ik\theta})$
$kLog_\pi (z) = k\left(\ln|re^{i\theta}| + iArg_\pi(re^{i\theta})\right) =k\ln (r) + ikArg_\pi(re^{i\theta}) $
Rasing $e$ to each power:
$e^{Log_\pi (z^k)} = e^{\ln|r^ke^{ik\theta}| + iArg_\pi(r^ke^{ik\theta})} = |r^ke^{ik\theta}|\cdot e^{iArg_\pi(r^ke^{ik\theta})}$
$e^{kLog_\pi (z)} = e^{k\ln (r) + ikArg_\pi(re^{i\theta})} = r^k \cdot e^{ikArg_\pi(re^{i\theta})} =r^k \cdot e^{ik\theta'}$ where $\theta + 2\pi n =\theta'\in(\pi, 3\pi], n\in \mathbb{Z}$
I don't know how to go from here. Can somebody please help?
Or can you just tell me under what condition the equation in title holds?
Edit: I asked my prof the same and his reply was:
This is almost never true. It is true only if $k$ is an integer and you use $log$ instead of $Log$. It is also true that the functions $αLog(z)$ and $Log(zα)$ differ by a constant.
But how did he conclude $(-i)^{i-1} = e^{(i-1)Log(-1)}$ like in his note above?
We often define the complex exponential [subject to the choice of branch, and using your collection of branches,] by $$z^w := \exp(w\operatorname{Log}_bz)$$ and as the definition of $\operatorname{Log}_b$, we also always have $z = \exp(\operatorname{Log}_bz)$. When you put those two properties together, you have for any complex $z,w$ (with $z$ nonzero), $$\exp(w\operatorname{Log}_bz) := z^w = \exp(\operatorname{Log}_b(z^w)).$$
When $w$ is an integer, this will be consistent with the algebraic definition since a integral multiple of an integral multiple of $2\pi$ will be an integral multiple of $2\pi$.
When $z>0$ and $w$ are real, then you need to choose the principal branch to find a consistent result to real exponentiation.
With the above definition in mind, the equation in your professor's notes, $$(-i)^{i-1} = e^{(i-1)\operatorname{Log}_b(-i)},$$ is just a trivial application of the definition of complex exponentiation.