What is essential non-isolated singularity, of $f(z) = \sin\left(\frac{1}{\cos(\frac{1}{z})}\right)$. Let a function $f$ be defined as $f(z) = \sin\left(\dfrac{1}{\cos\left(\frac{1}{z}\right)}\right)$.
I need to check what type of singularity it has at $z = 0$?
I found it is non-isolated as it is a limit point of the set of singularities.
Now in the answer key, It is written that it is essential and non-isolated.
Now this confuses me as I have read that essential singularity is a type of isolated singularity. I even googled and saw the same.
Can anyone throw some light on the meaning of essential non-isolated singularity?
It would be a great help.
 A: Let $f(z)$ be holomorphic on an annulus with laurent-expansion $$f(z)=\sum_{n\in\mathbb{Z}} a_n z^n \, .$$ $f(z)$ is said to have an essential singularity at $z=0$ if $$\limsup_{n\rightarrow -\infty} |a_{n}| \neq 0 \, .$$
The crucial point of an essential singularity is that $f(z)$ takes every value in $\mathbb{C}$ in the vicinity of $z=0$ (with at most 1 exception). Now if $f(z)$ takes every such value in the vicinity of $z=0$, then $1/f(z)$ takes every such value too (with said 1 exception). Therefore $1/f(z)$ has an essential singularity at $z=0$. Now if $g(z)$ is entire, its image covers $\mathbb{C}$ entirely and therefore $g(1/f(z))$ takes every value in $\mathbb{C}$ in the vicinity of $z=0$ (again the exception is passed on).
In your special case $g(z)=\sin(z)$ and $f(z)=\cos(1/z)$. It is non-isolated, since you actually have a multitude of such singularities in the vicinity of $z=0$ e.g. at $z_n=\frac{1}{\pi/2+n\pi}$ where $\cos(1/z)$ has a simple zero that is inside the $\sin$ (just like $\sin(1/z)$ is essential at $z=0$).
