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Could any one tell me with an example, what is the relation between Order and Multiplicity of a holomorphic function on Riemann Surface,and how this formuale comes?

for example let $p\in X$ be not a pole for $f$, and $f(p)=z_0$ then $f(z)-z_o$ has a simple zero at $p$ so $mult_p(f-f(p))=1$,$ord_p(f-f(p))=ord_p(f)-ord_p(f(p))=1+0?$

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    The formula $ord_p(f-g)=ord_p(f)-ord_p(g)$ that you seem to use is false2012-07-25

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Let me just quote Rick Miranda's "Algebraic Curves and Riemann Surfaces":

(Chapter II) Lemma 4.7: Let $f$ be a meromorphic function on a Riemann surface $X$, with associated holomorphic map $F\colon X \to \widehat{\mathbb{C}}$.

a) If $p \in X$ is a zero of $f$, then $\text{mult}_p(F) = \text{ord}_p(f)$.

b) If $p \in X$ is a pole of $f$, then $\text{mult}_p(F) = -\text{ord}_p(f)$.

c) If $p \in X$ neither a zero nor a pole of $f$, then $\text{mult}_p(F) = \text{ord}_p(f - f(p))$.

I don't have time right now, but perhaps later I'll include a proof and an example.

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    I find Miranda's trichotomy artificial: if $p$ is not a pole of $f$, then $\text{mult}_p(F) = \text{ord}_p(f - f(p))$, period. There is no need to artificially distinguish the cases $f(p)=0$ and $f(p)\neq 0$. In other words, I find it clearer to say that you have the dichotomy: either b) or else a)+c). But of course this is a matter of taste and Miranda's book is excellent anyway, so +1 for your mentioning it.2012-07-25