# 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 $(-2)^\pi$?
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I'd like to link this question with math.stackexchange.com/q/28511/8357 but I don't know how... I think there is some relevance. –  Pantelis Sopasakis Mar 28 '11 at 22:38

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.

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Let me mention that one can visualize the above Riemann surface in R^3 as an infinite number of copies of C which are cut along the same branch cut and glued together in an infinite staircase. Winding up and down this staircase is the topological analogue of analytic continuation clockwise and counterclockwise around the origin. –  Qiaochu Yuan Jul 30 '10 at 23:02

Wikipedia says "Complex powers of positive reals are defined via e^x as in section Complex powers of positive real numbers above. These are continuous functions. Trying to extend these functions to the general case of noninteger powers of complex numbers that are not positive reals leads to difficulties. Either we define discontinuous functions or multivalued functions."

So let me say a few more words. Essentially what we are looking for is $z^a=e^{a log(t)}$. But as you may or may not know, complex logarithms are problematic depending on your viewpoint.

In an answer to Harry's question, I describe a more careful way to define the complex exponential function, and suggest that one should just inverse this to get your logarithm. This way is safe and will yield your logarithm.

If you are less fickle, and are willing to take on the branched logarithms, may I suggest $Log(z)=ln\mid z\mid +iArg(z)$ where Arg is the principle argument. This will lead you to branched exponents.

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We can define complex powers as

$z^w := e^{(l(z)*w)}$ for $z,w \in \mathbb{C}$ and a complex logarithm function $l$

of course you need to make sure there actually exists such a logarithm $l:U\to\mathbb{C}$ function ($U$ needs to be a simply-connected subset of $\mathbb{C}\setminus\{0\}$

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Below is my geometric understanding of why irrational powers of negatives are difficult to define. As such it is probably not rigorous and may be wrong.

The way irrational powers of the real numbers are usually defined is by limits of fractional powers.

For complex numbers, the same is true, except the limiting process is more complicated. We can of course coast as usual on our real numbers result and see that we only need to define irrational powers on the unit circle, since every other points is some positive real multiple of a point on the unit circle.

Now in general z -> z^n wraps the circle around itself n times. What does z->z^(1/n) do? Well, it's not clear since each point has n possible points it could have come from, in particular if you partition the circle into n arcs of length 2pi/n, each of those gets mapped to the full circle. Once you choose a starting arc though, z -> z^m maps the starting arc to other arcs in the following way. Partition your starting arc of length 2pi/n into arcs of length 2pi/(nm) and then each of the little arcs gets mapped to a big arc that is 2pim/n away from the previous big arc.

The reason you choose arcs as opposed to weirdly distributed discrete sets, is because you want exponentiation to be continuous and hence you want the inverse image sets to be as the whole circle is connected.

There is no problem with fractional powers of -1, you have n choices for starting arc. But if you want your exponentiation to have some semblance of continuity with respect to the exponent, then you have to be choosing branches (arcs) for z->z^(1/n) that cohere, i.e. such that for very large n the roots get closer and closer together. This is done by requiring that the nth root of 1 is always 1, which makes all of the arcs into neighborhoods of 1. But this means that all the points with argument between 0 and pi get mapped in the half-arc above 1, and all the points with argument between pi and 2pi get mapped in the half-arc below 1.

Hence if you approach -1 from above and from below, the two limits of the nth root will be different, and hence you cannot have continuous exponentiation at -1.

As a result (you can do some more visualization if you wish), the irrational powers of -1 cannot even be defined as limits of rational powers.

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Note that all of the above is actually just a geometrization of the reason why the complex logarithm cannot be defined continuously on all of the unit circle. –  Vladimir Sotirov Aug 4 '10 at 21:15