Proposition

Show that if $f$ has an essential singularity at $z_0$ then $\frac{1}{f}$ also has an essential singularity at $z_0$.

Solution

Suppose $1/f$ does not have an essential singularity at $z_0$. Since $f$ has an isolated singularity at $z_0$, $1/f$ has an isolated singularity at $z_0$. By Proposition 9.5[a first course in complex analysis], $\lim_{z \rightarrow z_0} (z - z_0)^nf(z) \ne 0$ for any $n \in \mathbb{N}$. Moreover, this is a necessary and sufficient condition to having an essential singularity. Thus if $1/f$ does not have an essential singularity at $z_0$, then $\lim_{z \rightarrow z_0} (z - z_0)^n/f(z) = 0$ for some $n \in \mathbb{N}$.

Let $n \in \mathbb{N}$ be given. For any $r > 0$, there exists $z’$ in the punctured disk $D[z_0, r] \setminus \{ z_0 \}$ such that $\abs{0 - f(z)} < r^n$ by Casorati-Weierstrass. This implies that $\lim_{z \rightarrow z_0} (z - z_0)^n/f(z) \ne 0$. Therefore, $1/f$ has an essential singularity at $z_0$.