Do expressions like $(-1)^{2/3}$ show up naturally in pure or applied math? Background. Let $x$ denote an arbitrary real number. Then $x^n$ can be defined for each $n \in \mathbb{N}$ as follows:
$$x^n = \underbrace{x \times \cdots \times x}_n$$
If $x$ is furthermore non-zero, then we can extend the above definition by declaring that $x^k$ makes sense for each $k \in \mathbb{Z},$ by defining:
$$x^{-n} = \underbrace{(1/x) \times \cdots \times (1/x)}_n$$
for all positive integers $n$.
If $x$ is furthermore positive, we can go one step further and declare that $x^y$ makes sense for each $y \in \mathbb{R},$ by defining that
$$x^y = e^{y \log x}$$
for all non-integral real numbers $y$.
The result. The outcome of all this is rather pretty. We observe that all the usual exponential laws hold, and we get some nice closure results, too. In particular:


*

*if $n \in \mathbb{N}$ and $x$ is real, then $x^n$ is real.

*if $n \in \mathbb{Z}$ and $x$ is real non-zero, then $x^n$ is real-nonzero.

*if $y \in \mathbb{R}$ and $x$ is positive, then $x^y$ is positive.


The fourth step.  However, in high school we go one step further. For example, we learn that $(-1)^{1/3}$ is well-defined, and equals the unique real $x \in \mathbb{R}$ such that $x^3 = -1$. So too is $(-1)^{2/3},$ which equals $((-1)^2)^{1/3}$, or equivalently $((-1)^{1/3})^2.$
Unfortunately, once we make this fourth step, the theory suddenly becomes more complicated, because the usual exponential laws don't all hold. For example, $$-1 = (-1)^{2/2} = (-1)^{(1/2) \cdot 2} \neq ((-1)^{1/2})^{2} = \mathrm{undef}.$$

Question. Is this fourth and final step we learn in high school actually useful? More precisely, do expressions like $(-1)^{2/3}$ show up naturally in pure or applied math? If so, where?
If so, I'd like to see specific examples.

 A: Yes; in settings where a number $x$ has a unique $n$-th root, it can very well be useful to define $x^{1/n}$ to be that number.
The usual exponential laws do still hold if you don't try to take them too far; for example, if $x$ and $x^m$ both have unique $n$-th roots, then $(x^m)^{1/n} = (x^{1/n})^m$. In fact, we can say more: if $x$ has a unique $n$-th root, then so does $x^m$. And as a partial converse, if $x^m$ has a unique $n$-th root and $\gcd(m,n) = 1$, then $x$ has a unique $n$-th root as well.
I posit that in most purely real contexts where you would allow $x^3$ to be defined when $x$ is a negative real number, then you should also allow $x^{1/3}$ to be defined for negative real numbers.

I have recently made use of this, in the setting of $p$-adic analysis. Specifically, I was interested in the structure of $\mathbf{Z}_3[\sqrt{3}]$.
The power series for $\exp(z)$ is only defined for those elements such that $z \equiv 0 \pmod 3$; however, for this ring, it is convenient to extend the definition to all $z \equiv 0 \pmod{\sqrt{3}}$ by using $\exp(z) = \exp(3z)^{1/3}$.
