# Question about total branching number

Given a map $$f:\mathbb{CP^1} \leftarrow \mathbb{CP^1}$$ by $$f(z)=\frac{4z^2(z-1)^2}{(2z-1)^2}$$ Find all branching points and their degrees.

If my calculation is correct I got 4 branching points and in each point degree of $$f$$ is 2. Also I have concluded that inifnity is not a branching point. From Riemann Hurwitz formula I get that total branching number is 6 where degree of $$f$$ is 4. But if I directly compute total branching number I get 4. Where did I make a mistake?

$$f(z_0) = \infty$$ is in fact a branch point. There will be six ramification points $$z_0$$: $$\begin{matrix} z_0 & f(z_0) \\ \hline 0, 1 & 0 \\ \frac {1 \pm i} 2 & -1 \\ \frac 1 2, \infty & \infty \end{matrix},$$ each of index $$2$$. To see why this is the case, expand $$f(z)$$ around each $$z_0$$; the degree of the first non-constant term will be $$\pm 2$$.