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This is part of an exercise sheet in complex analysis. It should by solvable by rather elementary methods like the main theorems of complex analysis.

I succeded to show that $g$ has only finitely many zeros by using Bolzano-Weierstraß's theorem. If you divide $g$ by all its zeros you get a holomorphic function without zeros. I now fail to show that this new function is constant.

Peter
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Zardo
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    Let $g(z)=f(1/z)$. Then $g$ has what sort of singularity at the origin? (Removable singularity, pole, essential singularity.) That says what about the Laurent series for $g$? And then that says what about the power series for $f$? – David C. Ullrich Jun 25 '15 at 16:13
  • So I assume $e^{x^2}$ is not an entire function, then ? – Lucian Jun 25 '15 at 22:04
  • @Lucian $e^{z^2}$ is entire, but has an essential singularity at infinity. In other words, for any complex number $c$, you can find a path sequence $z_n$ tending to infinity, such that the $\lim_{n \to \infty} e^{z_n^2} = c$. – Steven Gubkin Jun 26 '15 at 01:48
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    How about: $f(z) = 1/g(z)$ is meromorphic on all of $\mathbb{C}$ with finitely many poles, and $\lim\limits_{z\to \infty} f(z) = 0$. For each pole $z_k$ of $f$, subtract the principal part $h_k$ of $f$ in $z_k$. Since the principal part contains only negative powers of $z-z_k$, we have $\lim\limits_{z\to\infty} h_k(z) = 0$. Then $$r(z) = f(z) - \sum_{k = 1}^n h_k(z)$$ is an entire function and $\lim\limits_{z\to\infty} r(z) = 0$. Hence $r\equiv 0$ by Liouville. Hence $f$ is rational, hence $g$ is rational, and an entire rational function is a polynomial. – Daniel Fischer Jun 26 '15 at 18:45

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Because it tends to infinity as $|z|$ tends to infinity, it is clear that the function has a pole at infinity. Looking at it on the riemann sphere, this means that it is a meromorphic function, with its only singularity at infinity. But you can prove that all meromorphic functions on the riemann sphere are rational functions, that is $P(z)/Q(z)$ where $P$ and $Q$ are polynomials. But since $Q$ has no zeroes in $\mathbb{C}$ (if it did, the function would not be entire) it must be constant by the fundamental theorem of algebra, so the function is equal to $P(z)$ and is a polynomial.

JHalliday
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  • That may be correct but does not fit the elementary methods I'm allowed to use here. It's a task my coaching student was given who had about two months of lectures in complex analysis by this point. – Zardo Jun 25 '15 at 22:57
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    Does he know about Taylor and Laurent series, particularly Taylor series at infinity? If so you can rephrase this by saying $g(z)=f(1/z)$ has a pole at 0, so its Laurent series has only finitely many negative terms. Because g is analytic everywhere except 0, you can say the principle part of the Laurent series agrees with g everywhere else, so everywhere except 0 g is a polynomial in $1/z$ , so f is clearly a polynomial in $z$. – JHalliday Jun 26 '15 at 00:41
  • Yes, this should work! :) – Zardo Jun 26 '15 at 13:06