how to determine deg f
$f(z)=\frac {z^{3}}{1-z^{2}} $
and what are its ramification points
The ramification point i found $p \in \mathbb{C}^{\infty}$ is a point with $multp(F) \geq 2$.
I found for example the following point:
$p=0$ because its a zero of f and so $multp(F)=ordp(F)=3 \geq 2$ The poles $+1$ and $−1$ are no ramification points sich their order are just $−1$. Also $\infty$ is no ramification points since
$f(\frac{1}{z}) = \frac{\frac{1}{z^3}}{1-\frac{1}{z^2}} = \frac{\frac{z^2}{z^2-1}}{z^3} = \frac{1}{z(z^2-1)}$ has a pole of order just 1 in zero (ord1(F)=−1).