In Griffiths' Introduction to Algebraic Curves. In the proof of the following statement,
Let $\omega$ be a meromorphic 1-form on a compact Riemann surface $C$, then $\sum_{p\in C} \operatorname{mult} _p(\omega) =-\chi(C) $.
The author showed that at a point $p$, in a suitable local coordinate, since we can write $\omega = z^n dz$, where $n$ is the multiplicity of $\omega$, then $\text{Re} (\omega) =r^n(\cos(-n\theta)\, dx+\sin(-n\theta)\, dy) $, so the index (which the author defined to be the winding number) $\text{Ind}_p(\text{Re} (\omega)) =-n$. Then it was concluded by Poincare Hopft Theorem that, $$\sum_{p\in C} \text{mult}_p(\omega) =-\sum_{p\in C} \text{Ind}_p(\text{Re} (\omega)) =-\chi(C) $$
Now comes my question, in the case where $\text{Re} (\omega)$ vanishes but $\omega$ does not vanish, the multiplicity (of $\omega$) is $0$ but the index may not be $0$. But for the formula above to hold true, the index should be $0$ at such point, how do we show it?
The proof in the book was correct. The reason why only considering the zeroes of $\operatorname{Re} (\omega) $ is sufficient, amounts to the fact that $$\operatorname{Re} (\omega)=0\iff \operatorname{Im} (\omega) =0\iff \omega=0$$
Fix a local coordinate $z$, we can express $\omega=f(z) dz$, for some meromorphic $f(z)=u(x, y) +iv(x, y) $. So $$\begin{align} \omega&=(u+iv) (dx+idy) \\&=(udx-vdy)+i(vdy+udy)\end{align}$$