Classification of singularities of $\sin\left( \frac{1}{\sin(\frac{1}{z})}\right)$

Question

I have to classify all singularities of the function $f(z) = \sin\left( \frac{1}{\sin(\frac{1}{z})}\right)$

I think the singularities appears on $\left\lbrace \begin{array}{ll} z=0 \\ z=\frac{1}{k\pi} \quad k \in \mathbb{Z} \setminus \{0\} \end{array}\right.$

But I don't know how classificates.

I tried to calculate the limit $$ \lim_{z\rightarrow 0} \ f(z)$$ but I can't prove that it doesn't exist. (If this limit doesn't exist, $z=0$ is essential singular point).

I tried calculate the Laurent expansion at $z=0$, but I don't know how to do it.

The same with another points $z=k \pi$

Anyone can help me? Thank you

Answer 1

1. $k\,\pi$, $k\in\Bbb Z$, are simple zeros of $\sin z$.
2. $1/(k\,\pi)$, $k\in\Bbb Z$, $k\ne0$, are simple zeros of $\sin (1/z)$; $z=0$ is an essential singularity of $\sin (1/z)$.
3. $1/(k\,\pi)$, $k\in\Bbb Z$, $k\ne0$, are simple poles of $1/(\sin (1/z))$.
4. $z=0$ is not an isolated singularity $1/(\sin (1/z))$, but a limit point of poles. The classification into avoidable, pole an essential applies only to isolated singularities.

Answer 2

https://en.wikipedia.org/wiki/Singularity_(mathematics)#Real_analysis

"The limits in this case are not infinite, but rather undefined: there is no value that $\sin\frac{1}{x}$ settles in on. Borrowing from complex analysis, this is sometimes called an essential singularity."

You have all the singularities down, and they are called essential singularities. To see what's happening, I recommend graphing it.