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So, I am trying to prove that if f(z) has a pole at infinity of order n then, its a polynomial. For that assume f(z) has a pole at infinity of order n. So, $f(\frac{1}{z})$ has pole at 0 of order n. So, we can write $f(\frac{1}{n}) = \frac{g(z)}{z^n}$, where g is nonzero and holomorphic around 0. We now have $g(z)=z^n f(\frac{1}{z})$. Consider $\frac{1}{z}$ and with this, we obtain $g(\frac{1}{w}) = \frac{f(w)}{w^n}$. I am trying to show that $g(\frac{1}{w})$ is bounded.

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  • $\begingroup$ Can you post a little more context? $\endgroup$ – Adam Hughes Dec 5 '16 at 20:44
  • $\begingroup$ So, I am trying to prove that if $f(z)$ has a pole at infinity of order n then, its a polynomial. For that assume $f(z)$ has a pole at infinity of order n. So, $f(\frac{1}{z})$ has pole at 0 of order n. So, we can write $f(\frac{1}{z}) = \frac{g(z)}{z^n}$, where g is nonzero and holomorphic around 0. We now have $g(z) = z^n f(\frac{1}{z})$. Consider $\frac{1}{z} = w$ and we get $g(\frac{1}{w}) = \frac{f(w)}{w^n}$. I am trying to show that $g(z)$ is bounded. $\endgroup$ – Avadutta Dec 5 '16 at 20:52
  • $\begingroup$ You should edit this into your question to avoid it being closed. $\endgroup$ – Adam Hughes Dec 5 '16 at 20:59
  • $\begingroup$ If you can prove that $g$ is bounded, you get that $f(z)=Cz^n$, which is not true. $\endgroup$ – N. S. Dec 5 '16 at 21:15
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Hint $$g^{(m)}(0)=\frac{n!}{2 \pi i}\int_{|z|=R} \frac{g(z)}{z^{m+1}} dz$$

Make $R \to \infty$ and try to express it as an integral in terms of $f$ around $0$. Prove that $g^{(m)}(0)=0$ for all $m >n$.

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