# continuous branch of an analytic function

For which real $\alpha,\beta>0$ is there a single valued branch $f$ of the analytic function $f(z)=z^{\alpha}(1-z)^{\beta}$ such that $f$ is defined on $\mathbb{C}\backslash[0,1]?$

I know that the argument of $f$ must not have any ambiguity when circumventing the points 0 and 1. Expressing $f$ in exponential form I get $$f(z)=e^{\alpha(ln|z|+iarg(z))+\beta(ln|1-z|+iarg(1-z))}$$.

I know $e^z$ is $2\pi i$ periodic, does this imply $2\pi\alpha+2\pi\beta=2\pi n$, where $n\in\mathbb{Z}?$

Yes, it means the condition is $\alpha + \beta \in \mathbb{Z}$.
One way to see it is to follow the change of the argument along small circles around the branch points, where we have a change of $2\pi \beta$ at $1$, and a change of $2\pi \alpha$ at $0$, and these must add to an integer multiple of $2\pi$ in order for the image of e.g. a dogbone contour to be a closed path.
Another way to see it is to look at the potential third branch point, $\infty$. Consider the function
$$g(z) = f\left(\frac{1}{z}\right) = z^{-\alpha} \left(1-z^{-1}\right)^\beta = z^{-(\alpha+\beta)}(z-1)^\beta$$
defined on $\mathbb{C}\setminus [1,+\infty]$. There is a holomorphic branch of $f$ on $\mathbb{C}\setminus [0,1]$ if and only if there is a meromorphic branch of $g$ on $\mathbb{C}\setminus [1,+\infty]$. The condition for the latter is easily seen to be $\alpha+\beta\in\mathbb{Z}$, so that $g$ has at worst a pole in $0$, and not a branch point.