Check the singularity of the function at $z = 0$

Question

Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$

1. a removale singularity

2. a pole

3. an essesntial singularity

4. a non isolated singularity

Since $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$ $$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$$

Thus $Sin\left(\frac{1}{1-y}\right) = Sin(1+y+y^2 +y^3+.......)$ = $\sum_{-\infty} ^{\infty}$$ a_n z^n$, thus $z=0$ is an issolated singularity.

Please check my solution is right or not. Also I want to know that how to check an non- isolated singularity

Answer 1

Take any neighborhood $U$ of $0$. Then there is a $k \in \mathbb{N}$ such that $\frac{1}{(2k+1)\pi}$ is in $U.$ Then $f(z)$ has a singularity at each of these points.