Is $i^i$ a single number or an infinite set of solutions? When I try to evaluate $i^i$, I get what I take to be a straightforward solution: $i^i = e^{-\frac{\pi}{2} + 2 \pi n}$ for $n \in \mathbb{Z}$. This amounts to basically a countably infinite set of real numbers.
However, everywhere I look, I see that $i^i = e^{-\frac{\pi}{2}}$ seems to be a well established identity.
When I think about it, it makes more sense to say that $i^i$ is a single value, rather than a set of infinite possible values. It's one thing to say that there are multiple solutions of $x$ to the equation $f(x) = 0$. So I could accept, for example, that the equation $x^{1/i} = i$ has infinite solutions. However, it's entirely different to say that a number can have infinite values. The latter statement seems straightforwardly wrong.
As far as I can tell, there are two possible ways to resolve this. Either $i^i$ isn't really a number, but rather, a set of numbers, or the function $f(x) = x^i$ is only defined for a specific branch of $x$, namely for $\arg(x) \in [-\pi, \pi)$ or something like that (similar to principal branch of Log).
Is one of the above the correct interpretation, or is there another resolution to this issue?
 A: Your intuition is pretty good. Think about how the equation $x^2 = 4$ has two solutions in the real numbers - $x = 2$ and $x = -2$. However, we define $\sqrt{4}$ to be just one of these solutions, specifically the positive one.
Similarly, in the complex numbers, the equation $z^n = w$ always has $n$ solutions when $n$ is an integer, and potentially infinitely many if it isn't. There are a few things we do to deal with this, including:

*

*Looking at Riemann surfaces, which are multiple copies of the complex plane "stitched together" via branch cuts; and


*Defining "principal values", i.e. picking one of the multiple solutions and declaring it to be the "standard" one in some sense, and then defining the other solutions in relation to it.
As for which one is correct - like so many things, it depends on context. Just like sometimes you're happy taking $\sqrt{4} = 2$ and that's all you need, and sometimes you need to keep both solutions and so you write $x = \pm 2$, even though that means that $x$ is somehow two numbers at the same time.
