Show that a function $f(z)$, which is analytic in the whole plane and has a non-essential singularity at $\infty$ reduces to a polynomial.

This question has been asked previously and I have gone through a solution. Following is a line of reasoning I came up with independently, but I need to know if it is sound enough.

$f(z)$ has no poles in the complex plane. Then $g(z) = f(\frac{1}{z})$ has a non-essential singularity at $z=0$ and none other. Suppose the algebraic order of this singularity of $g(z)$ is $h$. Following cases are possible.

  1. $h=0$. Then $g(0)\ne 0$ and $g(z)$ is analytic at $0$. Then $[f(z)]_{z\rightarrow \infty}$ is finite. Since $f(z)$ is analytic, from Liouville's Theorem it follows that $f(z)$ is a constant.
  2. $h\lt 0$. Then $g(z)$ has a zero of order $h$ and the same conclusion is possible as in 1.
  3. $h>0$. In this case $g(z)$ has a pole of order $h$ at $0$. By Taylor's Theorem $$z^hg(z)=B_h+B_{h-1}z+\cdots + B_1z^{h-1} + \phi(z)z^h$$where $\phi(z)$ is analytic at $0$. For $z\ne0$, it may be now written as $$g(z)=\frac{B_h}{z^h}+\cdots+\frac{B_1}{z}+\phi(z).$$Since the only singular part of $g(z)$ is in the RHS of above apart from $\phi(z)$, it must be true that $\phi(z)$ has no singularities in the extended plane. Thus $\phi(z)$ is a constant. By $f(z) = g(\frac{1}{z})$, $f(z)$ is a polynomial.


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    $\begingroup$ I like your proof but I have to wonder whether Taylor's theorem (in the complex form) comes "before" or "after" the theorem about entire functions that are well behaved at infinity being polynomials. If it comes after (that is, if it relies on the theorem you are proving) then using it to prove this theorem brings in circular reasoning. $\endgroup$ – Mark Fischler Jan 5 '16 at 0:26

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