Singularity of the function $f(z)=\sin\left(\cos\left(\frac{1}{z}\right)\right)$ at the point $z=0$ is ...

Question

For the function $f(z)=\sin(\cos(\frac{1}{z}))$, the point $z=0$ is

(a) a Removable singularity

(b) a pole

(c) an essential singularity

(d) Non-isolated singularity

I have written Laurent's series expansion $f(z)=\sin(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)= (1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)-\frac{(1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+...)^3}{3!}+...$ all the negative powers coming in the expansion. so singularity with respect to zero is an essential singularity. Am I correct? Please help me to judge.

Answer 1

No, it is not correct. For two reasons:

1. the Laurent series of $\cos\left(\frac1z\right)$ at $0$ is$$1-\frac1{2!z^2}+\frac1{4!z^4}-\cdots;$$
2. you did not prove that the negative powers do not cancel each other after a certain point.

You can prove that it is indeed an essential singularty using the Casorati-Weierstrass theorem: since $0$ is an essential singularity of $\cos\left(\frac1z\right)$, if $V$ is a neighborhood of $0$, then the set $W=\left\{\cos\left(\frac1z\right)\,\middle|\,z\in V\setminus\{0\}\right\}$ is a dense subset of $\mathbb C$ and, since $\sin$ is non-constant entire function, $\sin(W)$ is dense. Therefore, $0$ must be an essential singularity of your function.

Answer 2

Notice that the limit $\lim_{z \to 0} \cos(1/z)$ does not exist. Therefore the singularity is of essential type.