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This is an exercise from Alhfors Complex Analysis book- to show that an analytic function with a nonessential singularity at infinity must be a polynomial. It seems like it should probably be pretty straight forward, but I must be missing something. If it has a removable singularity at infinity then it extends to an analytic function on the Riemann sphere, and so must be constant by Liouville's theorem. What if there is a pole at infinity though? This was homework some time ago, and I never finished it :/ but have been thinking about it again recently. Thanks :)

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2 Answers 2

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Another hint: look at the function $f(\frac{1}{z})$ at z = 0, it has a nonessential singularity at 0...

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  • $\begingroup$ I think I have it now. Funny thing is I tried looking at f(1/z) before and for some reason missed it. I think I have it now. Thank you :) $\endgroup$
    – MJoszef
    Commented Mar 29, 2011 at 1:16
  • $\begingroup$ How do I write the proof formally? suppose f is not a polynomial. So positive power of $z$ is non-termnating. consider $f(1/z)$ The negative power is non-terminating. So has an essential singularity at $z=0$. So, $f$ has essential singularity at $f$. $\endgroup$
    – user464147
    Commented May 22, 2019 at 2:17
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Hint: consider the Laurent series in the annulus $0 < |z| < \infty$.

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  • $\begingroup$ Ahlfors book hasn't introduced Laurent series yet, so I was looking for a way along another path- hopefully. Thank you though. $\endgroup$
    – MJoszef
    Commented Mar 29, 2011 at 1:18

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