Limit point of poles is essential singularity? Am I speaking nonsense?

Question

The following is exercise 15 in section V.1 of Conway's Functions of One Complex Variable ("Classification of Singularities"). I'm currently studying for a complex analysis qualifying exam and this has appeared in the past.

Let $f$ be analytic in $G=\{z:0<|z-a|<r\}$ except that there is a sequence of poles $\{a_n\}$ in $G$ with $a_n\rightarrow a$. Show that for any $w$ in $\mathbb{C}$ there is a sequence $\{z_n\}$ in $G$ with $a=\lim z_n$ and $w=\lim f(z)$.

The conclusion makes me want to apply the Casorati-Weirstrass theorem. However, the singularity at $a$ is not isolated. As far as I know, an essential singularity is a particular type of isolated singularity. Am I wrong about this?

Any help would be greatly appreciated.

Answer 1

Hint: Suppose no such sequence exists.
Consider $g(z) = 1/(f(z) - w)$ (with $g(a_n) = 0$). What can you say about its singularity at $a$?

Answer 2

Define $V_\delta:=\{z\in G:|z-a|<\delta\}$. Fix $w$ in $\mathbb{C}$ and suppose that there is no such sequence. Then there exists an $\epsilon>0$ and $\delta>0$ such that $|f(z)-w|>\epsilon$ for all $z$ in $V_\delta\setminus\{a_n\}$. Now we can define the function $$g(z)=\frac{1}{f(z)-w},\quad z\in V_\delta$$ which is holomorphic throughout $V_\delta$ with zeros at each $a_n$.

Since $g$ is bounded by $1/\epsilon$, it follows that $g$ has a removable singularity at $a$. But since $a$ is a limit point for the zeros of $g$, the analytic continuation of $g$ to $V_\delta\cup\{a\}$ must have $g(a)=0$. Thus the set of zeros of the analytic continuation of $g$ has a limit point and therefore this analytic continuation must be identically zero. This contradicts the fact that $f$ is holomorphic in $G$ except at a sequence of poles. Therefore there must be such a sequence.