A negative base is a point of conflict between the three commonly used meanings of exponentiation.
- For the continuous real exponentiation operator, you're not allowed to have a negative base.
- For the discrete real exponentiation operator, we allow fractional exponents with odd denominators, and
$$(-a)^{b/c} = \sqrt[c]{(-a)^b}= \left( \sqrt[c]{-a} \right)^b = (-1)^b a^{b/c} $$
(and this is allowed because every real number has a unique $c$-th root)
- For the complex exponentiation operator, exponentiation is multivalued. An exponentiation with denominator $n$ generally takes on $n$ distinct values, although one is generally chosen as the "principal" value.
For $(-5)^{2/3}$, these three exponentiation operators give
- Undefined
- $\sqrt[3]{25}$
- $\omega \sqrt[3]{25}$ is the principal value. The other two are $\sqrt[3]{25}$ and $\omega^2 \sqrt[3]{25}$, where $\omega = -\frac{1}{2} + \mathbf{i} \frac{\sqrt{3}}{2}$ is a cube root of $1$.
Unfortunately, which meaning of exponentiation is meant is rarely ever stated explicitly, and has to be guessed from context.
I'm guessing that the second one is meant.
In case you're curious, here is part of the rationale for the first and third conventions.
In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other negative. And it's not even clear that $(-5)^{\pi/5}$ should be meaningful!
For the third convention, the whole thing is like the idea of $\pm 2$ being the 'square root of 4', but for the fact the complexes cannot be cleanly separated into "negative" and "positive" to let us choose a specific one nicely.
A method is chosen for the principal value, based trying to get positive bases 'right' and trying to keep continuity as much as possible, but alas this convention gets the negative bases 'wrong'.
In some sense, this can be viewed as the principal value of $(-5)^{2/3}$ chosen to be "two-thirds of the way" from positive to negative.
Best Answer
As other posters have indicated, the problem is that the complex logarithm isn't well-defined on $\mathbb{C}$. This is related to my comments in a recent question about the square root not being well-defined (since of course $\sqrt{z} = e^{ \frac{\log z}{2} }$).
One point of view is that the complex exponential $e^z : \mathbb{C} \to \mathbb{C}$ does not really have domain $\mathbb{C}$. Due to periodicity it really has domain $\mathbb{C}/2\pi i \mathbb{Z}$. So one way to define the complex logarithm is not as a function with range $\mathbb{C}$, but as a function with range $\mathbb{C}/2\pi i \mathbb{Z}$. Thus for example $\log 1 = 0, 2 \pi i, - 2 \pi i, ...$ and so forth.
So what are we doing when we don't do this? Well, let us suppose that for the time being we have decided that $\log 1 = 0$. This is how we get other values of the logarithm: using power series, we can define $\log (1 + z)$ for any $z$ with $|z| < 1$. We can now pick any number in this circle and take a power series expansion about that number to get a different power series whose circle of convergence is somewhere else. And by repeatedly changing the center of our power series, we can compute different values of the logarithm. This is called analytic continuation, and typically it proceeds by choosing a (say, smooth) path from $1$ to some other complex number and taking power series around different points in that path.
The problem you quickly run into is that the value of $\log z$ depends on the choice of path from $1$ to $z$. For example, the path $z = e^{2 \pi i t}, 0 \le t \le 1$ is a path from $1$ to $1$, and if you analytically continue the logarithm on it you will get $\log 1 = 2 \pi i$. And that is not what you wanted. (This is essentially the same as the contour integral of $\frac{1}{z}$ along this contour.)
One way around this problem is to arbitrarily choose a ray from the origin and declare that you are not allowed to analytically continue the logarithm through this ray. This is called choosing a branch cut, and it is not canonical, so I don't like it.
There is another way to resolve this situation, which is to consider the Riemann surface $(z, e^z) \subset \mathbb{C}^2$ and to think of the logarithm as the projection to the first coordinate from this surface to $\mathbb{C}$. So all the difficulties we have encountered above have been due to the fact that we have been trying to pretend that this projection has certain properties that it doesn't have. A closed path like $z = e^{2\pi i t}$ in which the logarithm starts and ends with different values corresponds to a path on this surface which starts and ends at different points, so there is no contradiction. This was Riemann's original motivation for defining Riemann surfaces, and it is this particular Riemann surface that powers things like the residue theorem.