Pole and limit point 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?
 A: 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.
A: 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 $G$.
