exponentiation - Non-integer powers of negative numbers

Roots behave strangely over complex numbers. Given this, how do non-integer powers behave over negative numbers? More specifically:

• Can we define fractional powers such as $(-2)^{-1.5}$?


• Can we define irrational powers $(-2)^\pi$?

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.