I'm facing the problem of proving that ($0,+\infty)$ is a branching point of $f(z)=z^z$. Now, straight from the definition, if $z=|z|e^{i\theta}$ with $0 \le \theta < 2\pi$ $ z^z := e^{z\log z} = \exp(z(\log |z| \, + \, i\arg z)) = \exp(z(\log |z| \, + \, i\theta \, + \, 2\pi im)) = $ $ = \exp(z(\log |z| + i\theta)) \exp(2\pi imz) $ The case with $z = \alpha$ (constant) is easily visualized by introducing a constant "phase factor" of $e^{2\pi i\alpha}$ by which the $e^{\alpha Log z}$ part is multiplied each time the function traverses a full circle, and so making a Riemann surface gluing together the e^{2\pi i\alpha m·}-branch with the $e^{2\pi i\alpha (m+1)}$-branch.
However, I cannot figure what to do with this, because the phase factor is not constant. How many branch point does this function have? How to prove that 0 (and $\infty$) is a branch point? And there's some clever way to do one or more branch cuts and construct a Riemann surface on which $z^z$ is continuous?
Thank you!