Problem: find the branch points of the function $$ f : \mathbb{C} \to \mathbb{P}^1 : z \mapsto \frac{1}{2}\bigg(z + \frac{1}{z}\bigg). $$
My try: The zeros are $i$ and $-i$, but I don't see why $f|V$ (the restriction of $f$ to $V$) is not injective for every neighbourhood $V$ of $\pm i$. I tried rewriting $f(z) = \frac{z^2 + 1}{2z}$ and viewing $f$ as a concatenation of maps, $$ \mathbb{C} \overset{z^2}{\longrightarrow} \mathbb{C} \overset{+1}{\longrightarrow} \mathbb{C} \overset{1/(2z)}{\longrightarrow} \mathbb{P}^1. $$ However, this did not help my understanding.
The map $f$ is injective in small neighbourhoods $V$ of $i$ or $-i$: that $f$ is zero at $\pm i$ is completely irrelevant.
The map $f$ will not be injective in small neighbourhoods of $a$ iff its derivative is zero there: $f'(a)=0$.
In your case the required two branching points are thus $a_1=1, a_2=-1$ .