Theorem. If $f$ has an essential singularity at $p$, then for any $a \in \mathbb{C}$ there exists a sequence $\{z_k\}$ such that:
\begin{equation*}
\lim_{n\rightarrow +\infty}z_n = p \quad \wedge \quad \lim_{n\rightarrow +\infty}f(z_n) = a
\end{equation*}
Proof I have so far. I believe that, so far I was able to prove the second condition, i.e., $\hspace{.2cm}\lim_{n\rightarrow +\infty}f(z_n) = a\hspace{.2cm}$ by contradition taking
\begin{equation*}
g(z) = \frac{1}{f(z)-a}
\end{equation*}
and then considering two cases $\overline{g}(p)\neq 0$ and $\overline{g}(p)=0$. I believe this part is somehow related to the Casorati-Weierstrass theorem.
The part where I am struggling is with the first condition, i.e.,
\begin{equation*}
\lim_{n\rightarrow +\infty}z_n = p
\end{equation*}
If someone could help me with this one, I would be really thankfull. (I understand this should be, somehow, related to the definition of convergence of a sequence).
Best Answer
Prove by contradiction. If the conclusion fails then there exists $r>0$ and $\delta >0$ such that $|f(z)-a| \geq \delta$ for $0<|z-p| <r$. Let $g=\frac 1 {f-a}$ Then $g$ is bounded near $p$ so it has a removable singularity at $p$. Obviously, $g$ is not identically $0$ in $D(p,r)$ so it has a zero of finite order (at most) at $p$. But then $f(z)=a+\frac 1g$ has a pole at $p$.