# Is there any relationship between ramification degree and order of a complex function?

Let $f$ be a map between compact Riemann surface.

Let $e_P(f)$ be the ramification degree of $f$ at point $P$.

Let $\nu_P(f)$ be the order of $f$ at point $P$, meaning that the lowest term of the Laurent series of $f$ has order $\nu_P(f)$.

I think there is some connection between this two. For example, I feel that when $\nu_P(f)>0$, $e_P(f)=\nu_P(f)$ and when $\nu_P(f)<0$, $e_P(f)=-\nu_P(f)$, am I right?

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The order of a map between Riemann surfaces isn't defined. Maps between Riemann surfaces are by definition holomorphic, in the sense that with respect to any choice of coordinates on the domain and codomain (near $P$ and $f(P)$), $f$ is a holomrphic function. –  Avi Steiner Feb 3 '13 at 3:31
I assume by ramification degree you mean multiplicity (what power of $n$ it locally looks like). The fact is that if $X$ is a Riemann surface and $f:X\to\mathbb{C}$ is meromorphic then $\text{mult}_p(F)=\text{ord}_p(f-f(p))$ if $p$ is not a pole of $f$ and $\text{mult}_p(F)=-\text{ord}_p(f)$ if $p$ is a pole of $f$ where $F$ is the induced map $\mathbb{C}\to\mathbb{P}^1$.
This is fairly easy to check. For example, by definition of multiplicity we can find a centered chart $(U,\varphi)$ at $p$ such that $\psi\circ F\circ\varphi^{-1}=z^{\text{mult}_p(F)}$ where $\psi$ is the standard centered chart at $\infty$ on $\mathbb{P}^1$. Now, the two functions $F\circ\varphi^{-1}$ and $f\circ\varphi^{-1}$ agree on a punctured neighborhood of $p$ and thus must have the same order at that point. Thus, $\text{ord}_p(f)=\text{ord}_p(F)$. But, $\psi$ has order $-1$ at infinity and so $\text{mult}_p(F)=\text{ord}_p(\psi\circ F\circ\varphi^{-1})=-\text{ord}_p(F)=\text{ord}_p(f)$.