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Let $f$ be holomorphic on $\mathbb{C}$. We have $f(z)=\sum_{n=0}^\infty a_nz^n$. Let $g$ be defined on $\mathbb C\setminus\{0\}$ by the Laurent series $g(z)=\sum_{n=0}^\infty \frac{a_n}{z^n}$. If $0$ is an essential singularity of $g$, then the coefficients satisfy $a_n \neq 0$ for infinitely many $n$, so we have an infinite number of terms in the Laurent series. Why does this imply that $f$ is not a polynomial?

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Corrected both mistakes. – Chris Jun 17 '12 at 21:20
up vote 2 down vote accepted

This comes down to the uniqueness of Taylor coefficients. By assumption, $f(z)=a_0+a_1z+a_2z^2+\cdots$, where the series converges on all of $\mathbb C$. If $f$ were a polynomial, then there would exist an $n\in\mathbb N$ and complex numbers $b_0,b_1,\ldots,b_n$ such that $f(z)=b_0+b_1z+b_2z^2+\cdots+b_nz^n$ for all $z\in\mathbb C$.

But then for $k\in\{0,1,2,\ldots,n\}$, $b_k=a_k=\dfrac{1}{k!}f^{(k)}(0)$, and for $k>n$, $a_k=\dfrac{1}{k!}f^{(k)}(0)=0$. Therefore, if $f$ is a polynomial, then $a_k=0$ for all but finitely many $k$.

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Thank you very much! – Chris Jun 17 '12 at 21:28

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