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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!

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$z^z$ is defined as $e^{z \log z}$, so you will have to have a branch of the logarithm to even define $z^z$. To see that the branch cut used in the definition doesn't dissappear:

Suppose you are using a branch of $\log z$ on a small neighborhood of some $z_0 \neq 0$ on the branch cut, and let $z_1$ be another point on the branch cut in that neighborhood. Then at $z_1$, the function $\log z$ jumps by $2\pi in$ for an integer $n$, which means $z \log z$ jumps by $2\pi i n z_1$ at $z_1$. As a result $e^{z \log z}$ will also jump at $z_1$ unless $n z_1$ is an integer, which will only occur at a countable set of points even taking the union over all $n$. Thus the branch cut remains; removing the countable set of points doesn't eliminate the branch cut.

So since $(0,\infty)$ is a branch cut for $\log (z)$ it is a branch cut for $z^z$, and since $0$ (and the point at infinity) is a branch point for $\log(z)$ it a branch point for $z^z$.