In the text I am currently reading through, I am not sure I understand correctly how they are defining a branch of $z^b$ where $b\in \mathbb{C}$ is fixed.

This is my understanding:

A branch of $z^b$ is a continuous function $g:G\rightarrow \mathbb{C}$ -- here $G$ is a region in $\mathbb{C}$ where there is a branch of $\log(z)$ -- such that $g(z) = \exp(bf(z))$ for some branch $f$ of $\log(z)$.

In other words the branches of $z^b$ are given by the continuous functions $$g(z) = \exp(b (\log|z| + i(\arg(z) +2\pi k))), k\in \mathbb{Z}$$

Is this resoning correct?


I'm not sure it makes sense to want "a banch" to be a particular thing. We can say that $z^{2/5}$ "has $5$ branches", but it is less meaningful to speak about a particular set of $5$ things that are the branches.

For example, here is a red dot and a shakily drawn curve that makes three windings around the dot:

red dot and shakily drawn curve

And we can easily agree that there are indeed three windings, but that doesn't mean that there's a definite answer to "what exactly is a winding" that makes the diagram contain exactly three instances of "a winding".

It is better to understand "a branch" to be a mostly informal and intuitive concept. If you're looking at the situation locally, with a particular simply-connected subset of the domain of the function in mind, then you can speak of branches of the function on that subset, and they will be well-defined things. But you cannot do the same for the entire domain, because they thing about branches is that they turn smoothly into each other as you go around a branch point.

  • $\begingroup$ So it makes sense to say: on a connected set $G$ the branches of $z^b$ are given by functions described in my post, while it makes less sense to try and describe a branch of a function without specifying the connected set $G$? $\endgroup$ – fosho Aug 28 '16 at 12:29

Some preliminary: given $z\in\Bbb C,\;\;z\neq0$, we define $\arg z$: it is called the principal argument and it's defined to be the unique $\theta\in]-\pi,\pi]$ such that $z=|z|e^{i\theta}$.

There are infinitely many different argument functions: once you have fixed $\theta_0\in\Bbb R$, you can define $$ \arg_{\theta_0}z:=\theta_0+\arg(ze^{-i\theta_0})\;\;. $$

The principal argument introduced at the beginning is the $\theta_0$-argument with $\theta_0=0$.

Now $z^b$ is defined by $$ z^b:=\exp\left[{{b(\log|z|+i(\arg_{\theta_0} z+2k\pi))}}\right] $$ with $\theta_0\in\Bbb R$ and $k\in\Bbb Z$, so the power function is defined up to a choice of the two parameter above $\theta_0$ and $k$, whose choice define the branch you are taking.

When $\theta_0=k=0$ you have the principal branch of $z^b$.

  • $\begingroup$ I thought the principle branch was when $k = 0$? That is the principle branch of $z^b$ uses the principle branch of the logarithm. $\endgroup$ – fosho Aug 28 '16 at 12:30
  • $\begingroup$ I just fixed Dman $\endgroup$ – Joe Aug 28 '16 at 12:32

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