# Branching points

1. Jul 21, 2015

### LagrangeEuler

$f(z)=z^{\frac{1}{2}}$ has brancing point at $z=0$.
$z=re^{i\varphi}=re^{i(\varphi+2n\pi)}$
From that
$$z^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i (\frac{\varphi}{2}+n\pi)}$$
For $n=0$, $z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}$
For $n=1$, $z^{\frac{1}{2}}=e^{i (\frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}$
For $n=2$, $z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}$
For $n=3$, $z^{\frac{1}{2}}=e^{i( \frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}$
For $n=4$, $z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}$
$...$
I see that this is multivalued function, but I am not sure why only problem or why only $z=0$ is branching point? Could you please explain me this?

2. Jul 21, 2015