Essential singularlites of the function $f(z)$ and $1/f(z)$? I read here that since $\sin(z)$ has an essential singularity at $\infty$, $1/\sin(z)$ also has an essential singularity at $\infty$. This got me thinking: Is it true that if $f(z)$ has an essential singularity at $z_0$ then $1/f(z)$ also has a singularity at $z_0$ and can this be proved in terms of Laurent series?

That $1/f(z)$ has a singularity can be shown by using Picard's theorem. We know that $f(z)$ takes all values except possibly one arbitrarily near $z_0$. This means that $f(z)$ will be arbitrarily close to $0$ arbitrarily near $z_0$ and therefor $1/f(z)$ is not bounded in any neighborhood of $z_0$.
