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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).

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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 $p$.

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  • $\begingroup$ I don't understand how this proves that \begin{equation*} \lim_{n\rightarrow +\infty}z_n = p \end{equation*} Isn't this a prove for what I have already shown? Thanks for your help $\endgroup$
    – xyz
    Jun 8, 2021 at 23:20
  • $\begingroup$ Try to write down the the negation of what you are asked to prove. That will give you existence of $r$ and $\delta$ as I have stated. @rpd $\endgroup$ Jun 8, 2021 at 23:23
  • $\begingroup$ I understand that. But aren't you proving (by contradition) that \begin{equation*} \lim_{n\rightarrow +\infty} f(z_n) = a \hspace{.1cm} ? \end{equation*} I wanted to prove what I've typed on my first comment, since I've already done this part. If I am missing something I am sorry! $\endgroup$
    – xyz
    Jun 8, 2021 at 23:25
  • $\begingroup$ Suppose what I stated in my first sentence is false. Take $r=\delta=\frac 1 n$. Then we get $z_n$ such that $0<|z_n-p| <\frac 1 n$ and $|f(z_n)-a| <\frac 1 n$. Doesn't that give both $z_n \to p$ and $f(z_n) \to a$? @rpd $\endgroup$ Jun 8, 2021 at 23:29
  • $\begingroup$ You are right indeed! Thanks for your help! $\endgroup$
    – xyz
    Jun 8, 2021 at 23:29

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