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This is an exercise problem from Complex Analysis by Joseph Bak and Donald J. Newman.

Suppose $f$ has an isolated singularity at $z_0$. Show that $z_0$ is an essential singularity if and only if there exist sequences $\{\alpha_n\}$ and $\{\beta_n\}$ with $$ \alpha_n \to z_0, \quad \beta_n \to z_0,\qquad\text{ and }\qquad f (\alpha_n) \to 0, \quad f (\beta_n) \to \infty.$$

I know Casorati–Weierstrass' theorem or Picard's theorem should guarantee that if $f$ has an essential singularity, then such suitable sequences should exist such that $f$ can approach any value, which we can set to be $0$ or $\infty$.

I, however, am not sure how to prove the converse. I'm wondering if anyone can offer a hint. Thanks.

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If it is a removable singlarity, then the limit $\lim\limits_{z\rightarrow z_0}f(z) $ exists.

If it is a pole, then the limit $\lim\limits_{z\rightarrow z_0}f(z) $ has to be infinite.(prove it)

Therefore, since none of the above happens, we have an essential singularity.

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When $f$ has a pole at $z_0\in{\mathbb C}$ then $\lim_{z\to z_0} f(z)=\infty\in\bar{\mathbb C}$. It follows that $\lim_{n\to\infty }f(\beta_n)=\infty$ for all sequences $(\beta_n)_{n\geq1}$ converging to $z_0$. In particular there is no sequence $\alpha_n\to z_0$ with $\lim_{n\to\infty}f(\alpha_n)=0$.

When $f$ has an essential singularity at $z_0$ then by the Casorati-Weierstrass theorem for each $n\geq1$ there is an $\alpha_n$ with $|\alpha_n-z_0|<{1\over n}$ and $|f(\alpha_n)|<{1\over n}$. It follows that $\lim_{n\to\infty}\alpha_n=z_0$ and $\lim_{n\to\infty} f(\alpha_n)=0$. Similarly, for each $n\geq 1$ there is a $\beta_n$ with $|\beta_n-z_0|<{1\over n}$ and $|f(\beta_n)|>n$. It follows that $\lim_{n\to\infty}\beta_n=z_0$ and $\lim_{n\to\infty} f(\beta_n)=\infty$.

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