Is $z^w$ for $z,w \in \mathbb{C}$ well-defined if you don't take a branch cut?

Question

If I understand correctly, $z^w$ for $z,w \in \mathbb{C}$ is a shorthand for $\exp(w\log(z))$. The problem is that every non-zero complex number has infinitely many logarithms. Say $\exp(y)=z$. Then, so is $\exp(y\pm2i\pi)$, $\exp(y\pm4i\pi)$, etc. Therefore, $z^w$ is a mutli-valued function. This is usually fixed by taking a branch cut, so that we require $\Im\left({\log(z)}\right) \in (-\pi,\pi]$, for instance. However, say we don't choose a branch of the logarithm. Is $z^w$ well-defined? Can I, for instance, write that $$ i^i=\{x \mid x=e^{-2 \pi n - \pi/2},n\in\mathbb{Z}\} \, ? $$ Another concern I have is about the notation $e^z$ being used in place of $\exp(z)$. I find this notation confusing, since $\exp$ denotes a single-valued function, whereas $e^z$ does not. If $z^w=\exp(w\log(z))$, then doesn't $$ e^z=\exp(w\log(e)) \, , $$ meaning that $e^z$ has infinitely many values?

Answer 1

You're right that every non-zero complex number has infinitely many logarithms, but it is possible that exponentiation will make all these values equal. For instance, let $ z= r e^{i \theta},$ then the possible values of $\log(z)$ are $ \log(r) + i \theta + 2 \pi i k$ for $k \in \mathbb{Z}.$ Now you would like $\exp(w\log z) $ to be a well-defined function on $\mathbb{C}$ i.e. you want all the possible values of the logarithm of $z$ to give the same value of $w^z.$ Writing this out, you get:

$$ \exp(w(\log r + i\theta + 2\pi i k )) = \exp ( w \log r + i\theta) \text{ for all } k \iff $$ $$ \exp( w \cdot 2\pi i k) = 1 \text{ for all } k \iff w \cdot 2\pi i k \in 2 \pi i \mathbb{Z} \text{ for all }k \iff w \in \mathbb{Z}.$$ Thus only exponentiating by integers is well defined without taking a branch cut, for example: $z^{1} $ is a well-defined analytic function.