# a question on meromorphic function at $\infty$

could any one help me to understand the Proposition $3.1$, specially why $p$ will be a polynomial? Thank you!

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Following the notation of the proof, first we construct $f(z) = p(z)/q(z)$, where $q$ is the polynomial whose zeroes are the poles of $f$ (with corresponding multiplicities), and thus we have $p(z) = f(z) q(z)$. Since both $f(z)$ and $q(z)$ are meromorphic on the Riemann sphere, then $p$ must also be meromorphic on the Riemann sphere.

However, $p$ is also a entire function on $\mathbb{C}$. So the key fact to prove is

The only entire functions on $\mathbb{C}$ that are meromorphic on $\hat{\mathbb{C}}$ are polynomials.

To prove this, write $p(z) = \sum_k a_k z^k$. Then $p(z)$ is entire on $\mathbb{C}$, so it only remains to check that $\infty$ is a pole. This is equivalent to showing that the function $g(z) : = p(1/z)$ has a pole at $z = 0$. The Laurent series expansion of $g(z)$ about $z = 0$ is precisely $g(z) = \sum_k a_k z^{-k}$. If indeed $g$ has a pole of order $N$ at $z = 0$, then $a_k = 0$ for all $k > N$ for some fixed $N$. But then $p(z)$ is a polynomial of order $N$.

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but you did not prove that there is a pole of order $N$ at $z=0$ of $g(z)$, you just assumed, am I right? –  La Belle Noiseuse Mar 7 '13 at 2:58
$g$ is meromorphic, so either it has a pole at zero or it is holomorphic at zero. –  Christopher A. Wong Mar 7 '13 at 7:57

Your function has a pole of finite order at $\infty$. Does this help?

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