# Sufficient condition for holomorphic branch

I am trying to solve the following question:

Find sufficient conditions on $a,b \in \mathbb{R}$ such that there exists an holomorphic branch of $z^a(1 - z)^b$ in the domain $\mathbb{C}\setminus[0,1]$

and I am not really sure how to approach it.

And I understand informally that our definition of $Log(z)$ and $z^\alpha$ on the complex plane is such that they are sets, and a branch is a continuous function into these sets (which means the branch cannot be defined on the entire plane). But I have no idea how to work with these definitions formally.

How do I prove that a branch exists (or does not exist) in a domain? What should be my conditions in this case?

Thanks! :)

Well first let's make things a bit easier for ourselves and think about what happens when $b=0$. Then we just have zed to the aye which is equal to $e^{a log(z)}$.
Well this is basically increasing the argument of $z$ by $2\pi$, so $log(z)$ increases by $2\pi i$. And so our function is multiplied by $e^{a\pi i}$. I'll pause and point out that this is the key observation. And for a holomorphic branch to exist, of course we must get the same value for our function both before and after we took our little walk.
Now the general case is no harder. You have two factors, but when going for a walk around the interval $[0,1]$, we're just multiplying our function now by $e^{a\pi i}e^{b\pi i}$, which we need to equal one, hence $a+b$ is an integer.
Then WLOG(!) $b=-a$, note that the image of $\mathbb{C}\setminus[0,1]$ under sending $z$ to $z/(1-z)$ is equal to the complex plane with the nonnegative real numbers removed, which is a set on which the logarithm can be consistently defined as an analytic function, so you're OK in this case.