Pole and limit point

Question

Question: If $f:G\longrightarrow \mathbb{C}$ is analytic except for poles show that the poles of $f$ cannot have a limit point in $G$.

My proof: Let $a_1, a_2, a_3, ..., a_k\in G; \forall k\in \mathbb{N}$ be is poles of $f$. By definition pole, we have that $\lim_{z \rightarrow a_k}|f|=\infty \mbox{ }\forall k\in\mathbb{N}$. Then, we have that for no pole of $f$ exist limit point, since your limit diverge for infinite. $\Box$

Is correct?

Possible duplicate of Limit point of poles is essential singularity? Am I speaking nonsense? — Alex M., Mar 18 '17 at 20:19
Also related: http://math.stackexchange.com/questions/1212296/accumulation-points-of-set-of-poles-for-meromorphic-functions — Alex M., Mar 18 '17 at 20:19
Your proof seems to say only finitley many poles. And your last sentence is incomprehensible. — zhw., Mar 18 '17 at 20:23

score 1 · Answer 1 · answered Apr 23 '19 at 13:22

Suppose the opposite is true i.e. there is a sequence of poles whose limit lies in G. Now consider the function 1/f. Then since earlier we had an infinite convergent sequence of poles now we instead have an infinite convergent sequence of zeroes. We now claim that the function 1/f must be the zero function which would then pose a contradiction for the existence of f.

For the claim simply observe that if f is analytic then even 1/f is. Now using the identity theorem for 1/f and the zero function say, 0, we get that 1/f = 0. There you have it!

To know about identity theorem click here

I am looking forward to any suggestions or corrections.

But $1/f$ is not analytic at zeros of $ f$? – William Leynoid Apr 20 '22 at 01:15 — William Leynoid, Apr 20 '22 at 01:15

score -1 · Answer 2 · answered Aug 07 '20 at 00:30

The identity theorem applies for analytic functions only so the reasoning above is inaccurate.

Instead, assume $z_0$ is a pole and a limit point in $G$ and there exist a sequence of poles $\{z_n\}$ such that $z_n \to z_0$ as $n \to \infty$.

$f(z)$ is analytic in the punctured disk $B(z_0, r)\setminus \{z_0\}$ as it is a meromorphic function. However, $B(z_0, r)\setminus \{z_0\}$ will intersect infinite poles in the sequence $\{z_n\}$ which is a contradiction. Therefore poles of $f$ cannot have a limit point in $G$.

Pole and limit point

2 Answers2