If $f: G \longrightarrow \mathbb C$ is analytic except for poles, then show that the poles of $f$ cannot have any limit point in $G$.
My attempt $:$
If the poles of $f$ has a limit point in $G$ then zeroes of $\dfrac {1} {f}$ has a limit point in $G$. Hence by identity theorem we have $\dfrac {1} {f} \equiv 0$ on $G$. But then there is no point in $G$ where $f$ can be analytic, a contradiction.
Is it the correct reasoning? If all the points of $G$ are poles of $f$ then vacuously the analyticity of $f$ follows. But then the above reasoning needs some modification. Isn't it? Please give me some suggestion regarding this.
Thank you in advance.
I think in Conway's book, poles should be isolated singularities. Hence the limit point of poles should not be a pole.
Let $a$ be the limit point of poles, then there is $a_n\rightarrow a$ where $a_n$ are poles. Since $f$ is analytic at $a$, we can find $r$ such that $f$ is analytic on $B_r(a)\subset G$. Pick $a_k\in B_r(a)-\{a\}$, and a sequence $\{z_n\}\subset B_r(a)$ such that $z_n\rightarrow a_k$. Then $\lim\limits_{n\rightarrow\infty} f(z_n) = f(a_k)\neq \infty$ contradicting that $a_k$ is a pole.