4

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?

  • The $e^{2\pi k\alpha i}$ in (1) is part of the definition of $z^{\alpha}$. You get that formula without that factor. Don't deduce it from $(-z)^\alpha=(-1)^\alpha z^\alpha$, which is not true in general. Deduce it from the definition $(-z)^{\alpha}=e^{\alpha\log(-z)}$. Put $z=re^{ix}$ and from this $-z=re^{ix+i\pi}$. – NDB Aug 04 '23 at 19:20
  • @NDB, in general, $$z=re^{ix}\Rightarrow -z=re^{ix\color{red}\pm i\pi}$$ as mentioned in this answer. – Matcha Latte Aug 04 '23 at 20:05
  • @Invisible That is not important, as the $\log$ already will output the $2\pi ki$. You could as well write $re^{ix+i\pi+2k\pi i}$, which is what one needs to do to compute $\log$. – NDB Aug 04 '23 at 20:07
  • @NDB Please provide counter example values $~z~$ and $~\alpha~$ such that $$(-z)^\alpha \neq (-1)^{\alpha} \times z^{\alpha} = \left[e^{i\pi}\right]^\alpha \times z^{\alpha}.$$ Here, in challenging your assertion, I am necessarily adopting the convention that Arg$(\alpha) \in (-\pi, \pi].$ – user2661923 Aug 04 '23 at 22:06
  • 3
    @user2661923 Oh you challenge? Have fun with $z=-1$ and $\alpha=i$. The convention on the argument of $\alpha$ is unimportant. The branch of the $\log$ is. Supposedly, you will use the same on both sides, on all factors. In any case, the main point is that, even if a particular instance of $(ab)^c=a^cb^c$ ends up being true, you don't use it in a deduction, because this is not true in general for all complex numbers. You could use the particular case, if you prove it first. Instead, use the definition of $a^b$ as $e^{b\log(a)}$, the properties of $e^x$ and the definition of $\log$. – NDB Aug 04 '23 at 23:01
  • There is not only a problem with the multivalued $(-z)^{\alpha}$. The same problem with $(zw)^\alpha$ which may or may not be $z^\alpha w^\alpha$ (all of them multivalued). Choosing a "principal value" for $\arg$ cannot always make $(zw)^\alpha = z^\alpha w^\alpha$. Frequent instances in this forum include questions about $\sqrt{(-1)(-1)} \ne \sqrt{-1}\sqrt{-1}$. – GEdgar Aug 05 '23 at 13:12

1 Answers1

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.

Gold
  • 28,141