Question: I was working on an analysis problem when this question arose in one my proofs. I think it may be either $-1$ or $1$, but it seems like there can only be an arbitrary way to assign this.

So is there an agreed upon method for determining $(-1)^x$ for irrational values of $x$?

Answer: 
[Euler's Formula](https://en.wikipedia.org/wiki/Euler%27s_formula) ($e^{ix}=\cos(x)+i\sin(x)$) is pretty useful here.
Since $\cos((2n+1)\pi)=-1$ and $\sin((2n+1)\pi)=0$ for integer $n$, $$\cos((2n+1)\pi)+i\sin((2n+1)\pi)=e^{i(2n+1)\pi}=-1$$

$$
(-1)^z = (e^{i(2n+1)\pi})^z = \cos((2n+1)\pi z)+i\sin((2n+1)\pi z) \tag{$n\in\mathbb{Z}$}
$$
(This equation can also be viewed as a [Generalization of De Moivre's formula](https://en.wikipedia.org/wiki/De_Moivre%27s_formula#Generalization))

Since $n$ goes over all integers, $(-1)^z$ has *infinite* values if the exponent $z$ is irrational. If $z$ is rational, its values repeat after a certain value of $n$.

Take $(-1)^\pi$ as an example:
$$
\begin{align}
(-1)^\pi &= (e^{i(2n+1)\pi})^\pi\\
&=e^{i(2n+1)\pi^2} \\
&=\cos((2n+1)\pi^2)+i\sin((2n+1)\pi^2) \tag{$n\in\mathbb{Z}$}
\end{align}
$$
The usual agreed-apon method, is to imagine that the imaginary part of the exponent is a line wrapped around the circle, the circumference is $2\pi$. The agreed on position is some point either in the range $-\pi\lt x \le\pi$ or $0 \le x \lt 2\pi$.

So if one has a number like $i$, then its complex logrithm is $\pi/2$. If we mean to raise it to some power, like $\pi^{17}$, then one calculates first $(\pi^{17}.\pi/2) \mod 2\pi$, that is, the remainder when an integer multiple of $2\pi$ is subtracted from the calculated value. And this remainder gives the power of the number raised to that power.

The solution for equations like $x^7=1$ is effected by $e^{2\pi ni}x^7=1$ for $n=0$ to $6$, and using the same model as above. When $n$ is not rational, this can go on forever, progressively splattering the unit circle with answers.