# Verify Herwitz's formula for $z^3/(1-z^2)$

This is an exercise from Miranda's book "Algebraic curves and Riemann surfaces".

Consider $$f(z)=z^3/(1-z^2)$$ as a holomorphic map from the Riemann sphere $$\mathbb{C}_\infty$$ to itself. Verify Herwitz's formula.

I found 1 zero and 3 poles.

1. Zeros: $$0$$, with an order and multiplicity of $$3$$.
2. Poles: $$1, -1$$ and $$\infty$$, with order $$-1$$ and multiplicity $$1$$.

These orders add up to $$0$$, which looks good. The degree of the map is $$3$$. Plug into the Herwitz's formula:

$$2g-2=\deg(F)(2g-2)+\sum_{p\in \mathbb{C}_\infty} [\text{mult}_p(F)-1]$$

And get $$-2=3\cdot(-2) + (3-1)$$ That is $$-2=-4$$. What did I do wrong?

• You need to look for critical values: where $f'(z)=0$. These correspond to ramification points. – Lord Shark the Unknown Jan 15 at 1:55
• Thanks I got it now. Besides 0, there are other 2 ramification points $\pm\sqrt{3}$ whose orders are 2. Hence they contribute another 2 to the right side and the equation holds. – Xipan Xiao Jan 15 at 7:28