Let $f$ be a meromorphic function on a Riemann surface $X$ , with associated holomorphic map $F:X\rightarrow \mathbb{C_{\infty}}$. Prove that :
a) If $p$ is zero of $f$ then $mult_p(F)=ord_p(f)$.
b)If $p$ is a pole of $f$ then $mult_p(F)=-ord_p(f)$.
This a lemma from Rick Miranda ' Algebraic curves and Riemann surfaces. I want to verify my idea:
a)Define $F(x)= f(x) $ if $x $ is not a pole
and $F(x)=\infty$ if $x$ is a pole.
We know that order of a meromorphic function at a point on a Riemann surface is invariant of charts. Now take two charts around $p$ , and $F(p)=f(p)=0$ say $(U,\phi)$ and $(V,I_{\mathbb{C}})$ respectively such that $I_{\mathbb{C}}\circ{F}\circ \phi^{-1}=F\circ \phi^{-1}$ is holomorphic and power series representation of $F\circ \phi^{-1}(z)-F\circ \phi^{-1}(\phi(p))=F\circ \phi^{-1}(z)-0=F\circ \phi^{-1}(z)$ around $\phi(p)$ has lowest degree say $m$ and $mult_p(F)=m$.
Now take take the same chart $(U,\phi)$ for $f$ at $p$ and $f\circ\phi^{-1}$ is has Laurent series expansion at $\phi(p)$ and $ord_p(f)>0$ , observe that $f\circ\phi^{-1}(z)= f(\phi^{-1}(z))=F(\phi^{-1}(z))\forall z\in\phi^{-1}(U) $ (I have doubt here).
therefore $Ord_p(f)=mult_p(F)$.
b) if $p$ is pole of $f$ then $F(p)=\infty$ and $p$ is a zero of $1/f$ now $ord_p(1/f)=-ord_p(f)$. And for $mult_p(F)$ we will take the chat around $p$ and $F(p)=\infty$ say $(U,\phi)$ and $(V,1/z)$ then power series of $\frac{1}{F\circ\phi^{-1}}$ is same as the Laurent series of $\frac{1}{f\circ\phi^{-1}}$ Therefore $mult_p(F)=ord_p(1/f)=-ord_p(f)$.
Hence the proof.