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This is based on Example 2.3 on Forster's "Lectures on Riemann Surfaces": let $f(z)=z^k+c_1z^{k-1}+...+c_k$ be a complex polynomial of degree $k$. Then $f$ can be considered as a holomorphic mapping $f:\mathbb{P}^1\rightarrow\mathbb{P}^1$ where $f(\infty )=\infty$. Using a chart about $\infty$, we can check that $\infty$ is taken with multiplicity $k$.

I am trying to understand this. Since $f$ is holomorphic at $\infty$, then if $w=\dfrac{1}{z}$, $f\left( \dfrac{1}{z}\right)$ is holomorphic in $0$. What I have is: $$ f(z)=z^k(1+c_1z^{-1}+...+c_kz^{-k}) $$ and $(1+c_1z^{-1}+...+c_kz^{-k})$ is holomorphic at $\infty$; therefore the multiplicity of $\infty$ is $k$.

Is this correct?

Marra
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  • $\mathbb P^1 $ is the projective space, right? – Dimitris Apr 21 '13 at 20:32
  • Yes, also noted by $\mathbb{CP}^1$. – Marra Apr 21 '13 at 20:33
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    Yes! Another way to see this is that the divisor associated to a (meromorphic) function on a (compact) Riemann surface always has degree $0$. When you throw out the point at infinity, $f$ has $k$ zeros and no poles by the Fundamental Theorem of Algebra. Thus all poles must happen at the point at infinity. Since the number of zeros must equal the number of poles, there must be a pole of order $k$ at infinity. – Matt Apr 21 '13 at 22:24
  • Thanks for the perfect explanation! I really appreciate it :) – Marra Apr 21 '13 at 22:31

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