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Let $f$ be holomorphic in all of $\mathbb{C}$ except for poles at $z=0$ and $z=1$. Further $f$ satisfies: \begin{equation} \displaystyle \lim_{|z|\to \infty} f(z)=0 \end{equation} I've to prove that $f$ is a rational function.

Approach: I considered the function $h(z) = z^n (z-1)^m f(z)$, with $n,m$ the orders of the poles $0$ and $1$. Clearly $h$ is holomorf in $\mathbb{C} \backslash \{0,1\}$. But because the limit of $h$ for $z\to 0$ or $z\to 1$ is finite (and non zero), by Riemanss theorem it follows that $h$ can be extended to an entire function. If $h$ has a removable singularity at $\infty$, we know that $h$ must be constant, so that $f$ is for sure a rational function. If I now can exlude the case that $\infty$ can be an essential singularity of $h$, we are done because then it follows that $h$ is polynomial and thus $f$ is a rational function. But how to approach this latter case? Thanks

bob
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    You don't need to split into cases. As $|z|\to \infty$ we have $|h(z)| \leq C |z|^{n+m}$ so $h$ is a polynomial by the extended Liovilles theorem. – Winther Apr 10 '17 at 10:29
  • Here is also a nice proof of that fact I think: http://math.stackexchange.com/questions/143468/entire-function-bounded-by-a-polynomial-is-a-polynomial (answer of John Adamski) – bob Apr 10 '17 at 11:43

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Hint: if you only had pole at $z=0$, you could do the following. Suppose $f(z)=\sum_{i=-n}^\infty a_iz^i$ near $0$. Then $g(z)=f(z)-\sum_{i=-n}^{-1}a_iz^i$ is holomorphic near $0$, and it is fairly easy to see that it is also holomorphic everywhere else, and limit at $\infty$ is $0$. By Liouville, $g=0$, whence $f=\sum_{i=-n}^{-1} a_iz^i$, which is a rational function.

Basically, the same thing can be done for $2$ (or any other finite number) of poles.

tomasz
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