How to calculate $\int_C \cos \big( \cos \frac 1z \big) dz$?

Question

The question says:

If $C$ is a closed curve enclosing origin in the positive sense. Then $\int_C \cos \big( \cos \frac 1z \big) dz=$ ?

$(1)\quad 0$

$(2)\quad 2\pi i$

$(3) \quad \pi i$

$(4)\quad -\pi i$

My thoughts :-

I am having trouble identifying the singularity type of $f(z)= \cos \big( \cos \frac 1z \big)$ at $z=0$. I know that $\cos (\frac 1z)$ has an essential singularity at zero. So is it valid to conclude that $f(z)$ also has essential singularity at zero by saying

" Since $\lim_{z\to 0}\cos (\frac 1z)$ does not exist , then $\lim_{z\to 0} f(z)$ also does not exist . "

Assuming I am correct, to calculate the integral, I just need to find the coefficient of $\frac 1z$ to apply Residue Theorem, Right ?

Now $f(z)=\cos \big( \cos (\frac 1z) \big)=1-\frac {\cos^2 (\frac 1z)}{2!} +\frac {\cos^4 (\frac 1z)}{4!}-... $

$=1-\frac 12 \big(1-\frac 1{2!z^2}+...\big)^2 + \frac 1{24}\big(1-\frac 1{2!z^2}+...\big)^4 $

So the coefficient of $\frac 1z$ is zero.

So my answer is $(1)$ , right ?

Any corrections or alternative ideas will be greatly appreciated. Thanks for your time.

Answer 1

(1) is the correct answer. But the sentence in quotation marks is not correct. You can say that $f(\frac 1 {2n\pi})=\cos 1$ and $f(\frac 1 {(2n+1)\pi/2})=\cos 0=1 \neq \cos 1$ so $f$ has an essential singularity since it is neither true that $f$ has a finite limit at $0$ nor is it true that $|f(z)| \to \infty$ as $z \to 0$. .

Answer 2

Your function is an even function. Therefore, the coefficient of $\frac1z$ is $0$, and so your integral is equal to $0$ too.