I understand that the function $f(z)=\sqrt{z}$ hast two branches and two branch points, ($z=0,z=\infty$). I also get that how more generally, $f(z)=\sqrt[n]{z}$ has the same branch points but has $n$ branches.
If we make them integers, things don't change. That is, $f(z)=\sqrt[n]{z}$, $n \in \mathbb{Z}$ has again the same branch points, and also $n$ different branches.
If $n \in \mathbb{Q}$ then $\sqrt[\frac{p}{q}]{z}=\sqrt[q]{(z^p)}$ which again has $q$ branches (here of course $p,q \in \mathbb{Z}$. But if we now let the discussed number be irrational, like $\pi$ for example, then I am not sure how many branches do we have. I mean, how many branches does this function have $f(z)=\sqrt[\pi]{z}$?
I am inclined to say it has an countably infinite number of them. I think this is true since $3,1414 \approx \pi$ and thus maybe $f(z)\approx\sqrt[31414]{z^{10000}}$ and this does have more than 30,000 branches. Is this nonsense?
And maybe, what could be said about the number of branches in the case where we consider even more weird stuff like $f(z)=\sqrt[i]{z}$ (where $i=\sqrt{-1}$ of course), $f(z)=\sqrt[\log(z)]{z}$ etc? Thanks in advance.
When you talk about $z^\alpha = e^{\alpha \log z}$ the problem is to define $\log z$.
If $z = re^{i\theta}$ then $\log z = \log r + i\theta$ so the problem reduces further to defining $\theta$. (The $\log r$ term denotes the real logarithm of the positive number $r$ which is perfectly well-defined and not a problem.)
The problem is that the value of $\theta$ is only defined up to adding a multiple of $2 \pi$. If you start at a point on the x-axis with $\theta=0$, and go around the origin once counterclockwise, then suddenly $\theta = 2\pi$ even though you have returned to your starting point.
Since $\theta$ is only defined up to adding a multiple of $2 \pi$, then $\log z$ is only defined up to adding a multiple of $2 \pi i$, so our original $e^{\alpha \log z}$ is only defined up to multiplying by a factor of $e^{2 \pi i k \alpha}$ for some integer $k$.
If $\alpha$ is an integer, then $e^{2 \pi i k \alpha} = 1$ for all integer $k$, so there is no problem: $z^{\alpha}$ is single-valued (only 1 branch) if $\alpha$ is an integer.
If $\alpha = p/q$ is a rational number, then $e^{2 \pi i k \alpha} = 1$ when $k$ is a multiple of $q$. You return to the original branch after $q$ loops around the origin. So you get $q$ branches for $z^{p/q}$, depending on $k$ modulo $q$.
If $\alpha$ is not a rational number, then $e^{2 \pi i k \alpha}$ never equals 1 (except for $k=0$) so you have infinitely many branches for $z^{\alpha}$, for each integer $k$. You never return to the original branch.