Determine singular point of $f(z)=\frac{e^z}{\sin\frac{1}{z}}$ at $z=0$ and $z=\infty$

Question

Determine singular point of $f(z)=\frac{e^z}{\sin\frac{1}{z}}$ at $z=0$ and $z=\infty$

I believe $z=0$ is an essential singular point because I cannot find the limit of this function and $z=\infty$ is removable singular point.

Am I right?

Answer 1

Since $e^z$ is non singular at $z=0$ and $e^0=1$, you can just look at $$ \frac{1}{\sin(1/z)} $$ and this has no limit for $z\to0$: approach $0$ with $\frac{2}{(2n+1)\pi}$ or $-\frac{2}{(2n+1)\pi}$.

The given function has no limit for $z\to\infty$ either. Consider instead $$ g(w)=\frac{e^{1/w}}{\sin w} $$ and its behavior at $0$. In order for $0$ to be a removable singularity you need $\lim\limits_{w\to0}e^{1/w}=0$, which doesn't hold. Can $w=0$ be a pole?