The question has been asked before, but I can't seem to wrap my head around it. I guess I'm missing something.

There is a certain part where I'm experiencing problems, and I can't seem to find an answer to that specific part in the other question(s).

If $f$ is entire and $\lim_{z\to \infty} f(z)= \infty$ then $f$ is polynomial.


Since $f$ is entire it can be expanded as a Taylor-series.

$$f(z) =\sum_{n=0}^{+\infty} a_n z^n$$

To prove $f$ would be polynomial I would have to prove $(\exists N)(\forall n > N)(a_n = 0)$


$$g(z) = f\left( \frac{1}{z}\right) = \sum_{n=0}^{+\infty} a_n\left(\frac{1}{z}\right)^n$$

Since $0$ is a pole (see below) of $g(z)$ the expansion above would be it's corresponding Laurent-expansion. And by definition of a pole $(\exists N)(\forall n >N)(a_n = 0)$.



Why is $0$ a pole of $g(z)$, why does $\lim_{z \to \infty} f(z) = \infty$ imply $0$ to be a pole of $g(z)$? I know $\lim_{z \to \infty} f(z) = \infty$ implies $\infty$ to be a pole of $f(z)$, but then...?


Well, $g(0)=f\left(\frac10\right)=f(\infty)=\infty$.

  • $\begingroup$ Hmm, how did I miss this :) Thanks $\endgroup$ – dietervdf Dec 24 '14 at 4:19
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    $\begingroup$ Sometimes one's thinking gets off track and it's easy to overlook the obvious. Often times you can catch it after putting the problem away for awhile and coming back to it with fresh eyes. :) $\endgroup$ – Tim Raczkowski Dec 24 '14 at 4:22
  • $\begingroup$ I think the author of this answer has essentially the right idea, the only problem is that $f(1/0)$ is technically meaningless. To be correct, you need to talk about $f(1/z)$ in arbitrarily small nrighborhoods of $0$. See Bombyx mori's answer and my comment! Cheers! $\endgroup$ – Robert Lewis Dec 24 '14 at 4:48
  • $\begingroup$ Yes I did wave my hands over the details, but it's not uncommon to view $1/z$ as a map of the extended complex plane (as realized by stereographic projection) into itself, and to say that this map sends 0 to the point at infinity. $\endgroup$ – Tim Raczkowski Dec 24 '14 at 4:59
  • $\begingroup$ OK, the point at $\infty$ passes, since 'tis the Yule. And who doesn't hand wave now and again? Holiday Cheers! $\endgroup$ – Robert Lewis Dec 24 '14 at 22:46

Note that when $z\rightarrow \infty, w=\frac{1}{z}\rightarrow 0$. And since $f(z)\rightarrow \infty$ as $z\rightarrow \infty$, there is no essential singularity. So $g(w)$ must have a pole of finite order at $\infty$.

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    $\begingroup$ Now that´s what I mean! Holiday Greetings! $\endgroup$ – Robert Lewis Dec 24 '14 at 4:48
  • $\begingroup$ @RobertLewis: Merry Christmas! $\endgroup$ – Bombyx mori Dec 24 '14 at 22:35

To show that 0 is a pole for f(1/z), try this:

Define $h(z)=\frac{1}{f(\frac{1}{z})}$, then $h(z)$ is analytic for $f(\frac{1}{z})$ non-zero. And note that $\lim_{z\to 0} |h(z)|=0 \implies lim_{z\to 0}h(z)=0$.

As the limit exists, we can take $h^*$ to be an analytic extension. Knowing that $h^*(z)$ is not identically zero, gives us that zeros are separated, so we can find a ball of radius $R>0$ such that $h^*(z)$ is analytic on $B_R(0)$.

Thus $h^*$ has a Taylor expansion centered about 0 on that region. As the first coefficient of that is 0 (Looking at the limit), but not all coefficients are non-zero or $h^*$ would be identically zero.

We have that $h^*(z)=z^N*\sum_{n=N}^\infty a_n z^{n-N}=z^N*k(z)$, where k is analytic and doesn't vanish on the disk.

$$lim_{z\to 0}Z^{N+1}f\left(\frac{1}{z}\right)=lim_{z\to 0} \frac{z}{k(z)}=0$$

Therefore $f(1/z)$ has a pole of order N at 0 $\implies$ f(z) has a pole of order N at infinity. Given this, dietervdf proof follows easily, but proving that the singularity at infinity is a pole and not an essential singularity is, I think, non-trivial.


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