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

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.

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.

• Thank you for pointing mistake. – user464147 Nov 5 '17 at 10:04
• Today I was trying to understand the proof. I understood everything but I am not able to see why $sin(W)$ being dense implies that $0$ must be an essential singularity of the function. Thanks in advance. – Aditya Prasad Jan 22 '19 at 16:30
• What are the alternatives? Well, $0$ could be a removable singularity. Or a pole. But in both cases, if you take a small neighborhood $U$ of $0$, then $f\bigl(U\setminus\{0\}\bigr)$ would not be dense in $\mathbb C$. Since it is… – José Carlos Santos Jan 22 '19 at 16:44
• ...it is a isolated singularity. Am I right? – Aditya Prasad Jan 22 '19 at 17:21
• That should be obvious. No: it is an essential singularity. – José Carlos Santos Jan 22 '19 at 17:24

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

• consider $\frac{1}{z}$limit at $0$ , doesn't exists. But it is pole, right? – user464147 Nov 5 '17 at 10:09
• @Maneesh_Narayanan If a point $z = z_0$ is a pole of $f (z)$, then $\lim_{z \to z_0} f (z) = \infty$. – user371838 Nov 5 '17 at 10:37
• Ok. I got the point. Thank You. Is that enough to show that limit at that point is neither 0 or infinity and limit doesn't exists, in order to prove some point is an essential sigularity? – user464147 Nov 5 '17 at 10:42
• math.stackexchange.com/questions/1806582/… – user464147 Nov 5 '17 at 10:46
• this one is essential right? – user464147 Nov 5 '17 at 10:47