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

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?

• Mathematicians normally assume $e^z$ is short for $\exp z$, thereby taking the $\log e=1$ branch. Others are disinteresting, or we wouldn't be using the base $e$.
– J.G.
Commented Jan 1, 2021 at 14:20
• @J.G. Thank you, this does seem to clear up my confusion with $e^z$ versus $\exp(z)$. Does $z^w$ even make sense if we don't specify what branch of the logarithm we are using?
– Joe
Commented Jan 1, 2021 at 14:23
• It does need to be specified, at least by context (a lot of complex analysis boils down to carefully considering whether a contour jumps branches) or convention.
– J.G.
Commented Jan 1, 2021 at 15:04
• @J.G. In your experience, do people ever define $z^w$ as a multifunction? For example, do you ever see $$z^w=\{x : x = \exp(wr), \text{ where r is a logarithm of z}\} \, ?$$And would this definition be useful?
– Joe
Commented Jan 2, 2021 at 16:47
• @J.G. If you have the time to respond to my query, then I would be very grateful. But thanks anyway.
– Joe
Commented Jan 5, 2021 at 15:40

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.
• Thank you for this answer. I didn't know that only exponentiating by integers is well-defined (unless you take a branch cut). However, I'm still unsure about this: is it correct to write $$i^i=\{x \mid x=e^{-2 \pi n - \pi/2},n\in\mathbb{Z}\} \, ?$$
• I would take that to mean the set of all the possible values of $i^i$ defined by choosing a branch of the logarithm. In practice, however, you define a branch of $z^i$, then look at the value of this branch at $i,$ which is one of the values in your set. You can, however, define a "multi-valued" map: $\mathbb{C} \to \mathcal{P}(\mathbb{C}), z \mapsto$ $\{$ set of possible values...$\},$ but I have absolutely no idea how to deal with such object: What does mean for it to be continuous, analytic, etc? But we can answer these questions if we choose a branch and work with normal maps. Commented Jan 1, 2021 at 17:05