Is "the nth root of x" well-defined without further qualification?

Question

(No, I'm not asking if $\sqrt{-1} = +i$ or if $\sqrt{-1} = -i$. Yes, I know $+i$ is the principal square root.)

Consider the cube root of -8. If asked to evaluate it, I would say -2, and I think we all agree.
But if $r \in \mathbb{R}$ and $0 \leq r$, $0 \leq \theta < 2\pi$ and I'm asked to evaluate the cube root of $r e^{i\theta}$, I would say $\sqrt[3]{r} e^{i\theta/3}$. Yet that would imply the cube root of $-8$ would be $1+i\sqrt{3}$, not $-2$.

Basically, it seems to me that a number is more than just its value. Its "type" (I'm using that word from a computer science standpoint), which in this case is real vs. complex, seems to affect its meaning. But I don't remember ever learning that I need to distinguish between "real numbers" and "complex numbers that happen to be real". $-8$ was always just... $-8$.

Is there a nice way to resolve this that I'm missing? For example, is there a canonical definition for "the cube root of $-8$" that doesn't require saying whether $-8$ is real or complex? Or do mathematicians make a distinctions between the "type" of a variable when evaluating it, as opposed to merely its value? Or something else?

Answer 1

"The" cube root implies there is only one. $z^n = c$ has $n$ solutions up to multiplicity (i.e. any complex number can have up to 3 different cube roots).

That being said, the "type" of the number in this case is which field you're using, which is part of the expression you're writing.

Answer 2

-2 is the "principal cube root", which is the unique real solution to $y^3 = x$ (the only complex solution with argument 0 or $\pi$).

$1 + i\sqrt{3}$ is the "principal complex cube root", which is the complex solution with the smallest non-negative argument.

Naturally the two are going to differ for negative $x$ for any odd power. So yes, you just have to specify whether you're interested in real or complex answers. And yes, that does mean that, practically speaking, you will treat a complex number that "happens to be real" differently from a real number, because if you're working strictly with reals you will probably find the principal root the most useful, and if you're working with complex numbers in general, you will probably have a lot more use for the principal complex root, and it would be weird and inconvenient to have a special case for numbers without an imaginary part.