Let $S$ be a compact oriented surface, given a smooth real 1-form $\alpha$ which vanishing at disrete set $\Delta$.under local coordiante $$\alpha = \alpha_1 dx+ \alpha_2 dy$$
Define the multiplicity of $\alpha$ at $p\in \Delta$ be the winding number of the map $$(\alpha_1,\alpha_2):S^1(r) \to \Bbb{R}^2\setminus \{0\}$$ where $S^1(r)$ is some small circle around the disrecte zero point $p$ of radius $r$
It can be shown using the complex anaysis result that for a holomorphic 1-form $\alpha = fdz $ on a compact Riemann surface $S$ the multiplicity of zeros equals to the multiplicity for the real 1-form $\alpha +\bar{\alpha}$ defined above.(therefore the sum of multiplicity of $f$ equals to $-\chi(S)$)
The question is how can I extend this argument to meromorphic 1-form
In Donaldson's Riemann surface book, he did it as follows,which seems to contruct some surrogate holomorphic 1-form for the meromorphic 1-form:
We can extend this discussion to meromorphic 1-forms. To do this, we fix an area form $\omega$ on $X$. This means that we can define a Hermitian metric on $T^{*} X^{\prime}$ : $$ \xi \wedge \bar{\xi}=|\xi|^{2} \omega . $$ Suppose $\alpha$ is a meromorphic 1 -form on $X$. Choose a real-valued function $p$ on $\mathbf{R}$ with $p(t)=1$ for small $t$ and $p(t)=t^{-1}$ for large $t$. Now define $$ \tilde{\alpha}=p\left(|\alpha|^{2}\right) \alpha $$ away from the poles of $\alpha$, and $\tilde{\alpha}=0$ at the poles of $\alpha$. Locally, around a pole of $\alpha$, we have $$ \tilde{\alpha}=\frac{1}{|f(z)|^{2}} f(z) R d z=\frac{1}{\overline{f(z)}} R d z, $$ where $R$ is a smooth strictly positive function, determined by the area form $\omega=R d x \wedge d y$. Thus $\tilde{\alpha}$ is smooth and its zero set is the union of the zeros and poles of $\alpha$. It is clear that the zeros of $\tilde{\alpha}$ corresponding to the poles of $\alpha$ have a multiplicity equal to minus the order of the pole. Thus we have the multiplicity formula for the meromorphic 1-form.
However, I can't get the idea how $\tilde{\alpha}$ looks like?
The goal of $p(t) = 1/t$ is to beat the growth of $\alpha$ near the pole.As the expression $p(|\alpha|^2)\alpha$ near the pole shows.Therefore eliminate all the singular points.
The key point is to compute the multiplicity of $\tilde{\alpha}$.At the zeros of $\tilde{\alpha}$ correspond to the zeros of $\alpha$, it has same Taylor expansion near the zero point,since they coinside inside a small neighborhood near the zero points.therefore same multiplicity.
For the zeros of $\tilde{\alpha}$ correspond to the pole of $\alpha = fdz$, the Laurant expansion of $f$ locally looks like $$f(z) =a_{-n}z^{-n}+ ...+a_0 +a_1z+a_2z^2... = z^{-n}g(z)$$ where $g(z)$ is some nonvanishing holomorphic function, therefore $$\frac{z^{-n}}{|z|^{-2n}} \overline{g(z)} = \overline{z^ng(z)}$$ as conjugation of a holomorphic function has mutiplicity $-n$.
Remark: it can be checked the conjugate of the holomorphic 1-form has minus the multiplicity of the original holomorphic 1-form.(with both multiplicity defined using the real 1-form $\alpha +\bar{\alpha}$).