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I am reading the text on Riemann surfaces by Farkas and Kra, and I'm having trouble understanding the following step in the Proposition on p. 12.

...it follows (even if the points $P_j$ are not distinct) that $\sum \limits_{P \in f^{-1}(Q)} (b_f(P) + 1) \geq n$.

Clearly, this is true if the $P_j$ are all distinct. However, I do not see the case when a group of say, $k$ of them are not distinct. I would need to show that the ramification number of $Q$ is $m$. In coordinates, this reduces to the following, which I don't see how to show.

Let $f: \mathbb{C} \to \mathbb{C}$ be analytic in a neighborhood of zero. Let $\{z_n\}$ be a sequence of complex numbers converging to zero. For each $n$, let $p_n^1, \ldots, p_n^k$ be distinct complex numbers such that $f(p_n^j) = z_n$, and $\lim \limits_{n \to \infty} z_n^j = 0$ for all $1 \leq j \leq k$. Then $f$ has a zero of mutiplicity $k$ at $z= 0$.

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You should specify notation! What's $b_f$, for example? –  Dylan Moreland Nov 19 '11 at 21:24

1 Answer 1

Use the argument principle --- take a small circle $\gamma$ around $0$ so that $f$ has no other roots in the closure of the circle apart from $0$.

For $k$ sufficiently large, the number of points (counting multiplicity) inside the disk bounded by $\gamma$ where $f$ takes the value $z$ is given by a certain path integral along $\gamma$ that depends continuously on $z$ ...

(Alternatively, note that by composing with a suitable choice of coordinate, $f$ can be brought into the form $z\mapsto z^m$, where $m$ is the multiplicity of $0$.)

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