Let the function $$ f(z) = \frac{1}{\sin(1/z)}, \quad z \in \mathbb{C} $$
I've been able to find the singularities. Now I would like the show that these are simple poles and also show that $z = 0$ is an essential singularity.
First, I'm looking for the zeros of $\sin(1/z)$.
\begin{align} \sin(1/z) = \frac{e^{i/z} - e^{-(i/z)}}{2i} = 0 &\iff e^{iz} = e^{-(i/z)} \\ &\iff e^{2i/z} = 1 \\ &\iff e^{2i/z} = - e^{i\pi} \\ &\iff \frac{2i}{z} = i\pi + i2\pi k, \quad k \in \mathbb{Z} \\ &\iff z_k = \frac{1}{(\pi/2) + \pi k}, \quad k \in \mathbb{Z} \end{align}
We also see that the function $z \rightarrow \frac{1}{z}$ is holomorphic on $\mathbb{C} \setminus \{0\}$
Using the properties of compositions and inverse of holomorphic function. $f$ is holomorphic on $\mathbb{C} \setminus \{ 0, z_0, \dots, z_m \}, \quad m \in \mathbb{Z}$
Now, I guess $z_0, \dots, z_m$ are simple poles since they are neither removable singularity or essential singularity. To show that I should calculate for every $z_k$, $$ \lim_{z \rightarrow z_0} (z - z_0) f(z) $$ which is also the coefficient $a_{-1}$ of the Laurent serie. If these are simple poles we should have $a_{-1} \neq 0$.
For example, for $z_0 = \frac{2}{\pi}$ we have, $$ g(z) = \left(z - \frac{2}{\pi} \right) \frac{1}{\sin(1/z)} $$
But the limit, $$ \lim_{z \rightarrow \frac{2}{\pi}} g(z) = 0 $$ right ? Seems to be the same for the other $z_k$. So that doesn't work. How to show these are simples poles ?
Eventually, the point $z = 0$ is an essential singularity. To show that I would like to find a path where the limit is different. I've been trying to use the taylor serie of $\sin(1/z)$.
$$ \sin(1/z) = \sum_{n=0}^{+\infty} (-1)^n \frac{1}{z^{2n+1}(2n+1)!}, \quad \forall z \in \mathbb{C} $$
But didn't find anything conclusive. Is there any other way to show that for $z = 0$ ?
Note that\begin{align}\sin\left(\frac1z\right)=0&\iff\frac1z=\pm k\pi\text{ for some }k\in\Bbb N\\&\iff z=\pm\frac1{k\pi}\text{ for some }k\in\Bbb N.\end{align}And those points are simples poles of $f$, since they are simple zeros of $\sin\left(\frac1z\right)$.
On the other hand, this shows that $0$ is not an isolated singularity. Therefore, it makes no sense to assert that it is (or that it is not) an isolated singulary (or a pole or a removable singularity).