Writing $(-z)^\alpha$ in terms of $z^\alpha$

Question

I have a very silly confusion in complex analysis. Let $z\in \mathbb{C}$ and $\alpha\in \mathbb{C}$. Now consider $(-z)^\alpha$ and suppose that we want to relate this to $z^\alpha$. One thing that I immediately notice is that $(-z)^{\alpha} = (-1)^\alpha z^\alpha$ seems to be ambiguous.

In fact, we can write $-1 = e^{i\pi+2\pi k i}$ for $k\in \mathbb{Z}$, but the choices seem to give different results depending on what $\alpha$ is. In fact we have

$$(-z)^\alpha = e^{2\pi k\alpha i} e^{i\alpha \pi} z^\alpha\tag{1}.$$

Now the prefactor $e^{2\pi k\alpha i}$ is in general non-trivial if $\alpha\notin \mathbb{Z}$.

I feel that the right way to write $(-z)^\alpha$ in terms of $z^\alpha$ is by taking into account the branch cut, but I feel a bit confused in how this should be done correctly.

So what is going on here? Why writing $(-z)^\alpha$ in terms of $z^\alpha$ seems highly ambiguous? How to identify the correct choice in a given situation?

Answer 1

The definition of exponentiation is

$$z^\alpha = \exp\left[\alpha \log z\right]\tag{1}$$

and in turn the definition of $\log z$ is

$$\log z=\ln|z|+i\arg(z)\tag{2}.$$

With this we can already see wherein lies all the issues, namely, in the definition of $\arg z$. To fully define $\log z$, and hence $z^\alpha$ we need to choose a range for $\arg z$. Let's say we do this by specifying one initial angle $\theta_0$ and demanding that $\arg(z)\in (\theta_0,\theta_0+2\pi)$. Once this is done the question can be given a precise answer.

To do so we calculate

$$(-z)^\alpha = \exp\left[\alpha\log(-z)\right]=\exp\left[\alpha \ln |-z|+i\arg(-z)\right]\tag{3}.$$

The thing now is to relate $\arg(-z)$ to $\arg(z)$. If you draw this on the complex plane you'll notice that geometrically it is clear that, having in mind keep $\arg(z)\in (\theta_0,\theta_0+2\pi)$, there are only two possibilities for $\arg(-z)$, namely $\arg(-z)=\arg(z)\pm \pi$.But which has to be chosen?

Well, it is simple. We define it so that $\arg(-z)\in (\theta_0,\theta_0+2\pi)$. The line defined by $\theta_0$ splits the complex plane in two half-planes. I'll call lower half-plane the one described by $\theta\in(\theta_0,\theta_0+\pi)$ and upper half-plane the one described by $(\theta_0+\pi,\theta_0+2\pi)$. It is then easy to see that

1. If $-z$ is in the lower half-plane, then $\arg(-z)=\arg(z)-\pi$; In this case we have $$(-z)^\alpha = \exp\left[\alpha \ln|z|+i\arg(z)-i\pi\alpha\right]=e^{-i\pi\alpha}z^\alpha$$

2. If $-z$ is in the upper half-plane, then $\arg(-z)=\arg(z)+\pi$; In this case we have $$(-z)^\alpha = \exp\left[\alpha\ln|z|+i\arg(z)+i\pi\alpha\right]=e^{i\pi\alpha}z^\alpha$$

Observe that what determines this is therefore at which side of the branch cut $-z$ is.