Residue Theorem Integral of $\cos(e^z)/\sin^2(z)$

Question

I need help with the following integral: $$ \frac{1}{2\pi i }\int_{|z| = 4} \frac{\cos(e^z)}{\sin^2(z)} dz. $$ I have tried finding the residues, which gave me 0 through my (possibly erroneous) calculations, making the power reduction substitution $\sin^2(z) = 1/2 (1- \cos(2z))$, and parameterizing the curve via $\gamma(t) = 4e^{it}$ and going from there without any luck. Any hints or nudges in the right direction would be cherished. Thanks!

Edit: for context, this integral was the first problem on a qualifying exam in my department, and I have been stuck on it in preparation for the next one. It is therefore important that I can find a solution that I can do by hand in a timely manner- the residue calculations I have tried are extremely cumbersome and haven't produced anything useful yet.

Answer 1

To elaborate on Nitin's answer, you were on the right track with residues, but the actual calculation is tricky. The key fact is $ \lim\limits_{z\to 0}\left[\,\frac{\sin(z)}{z}\,\right] = 1 $, which can be applied because the limit of products is the product of limits.

By the product rule, $ \lim\limits_{z\to 0}[\,z^2 f(z)\,]' = \lim\limits_{z\to 0}[\,2z\, f(z) + z^2 f'(z)] $.

By the quotient rule, $f'(z)= \dfrac{-e^z \sin(e^z) \sin^2(z) - 2\cos(e^z) \cos(z) \sin(z)}{\sin^4(z)} $

Then $\text{Res}(f,0) = \lim\limits_{z\to 0}\left[\, 2z\dfrac{\cos(e^z)}{\sin^2(z)} + z^2\dfrac{-e^z \sin(e^z) \sin^2(z) - 2\cos(e^z) \cos(z) \sin(z)}{\sin^4(z)}\,\right] $

Focusing in on the expression inside the limit, I will make some rearrangements

$$ \left(\frac{z}{\sin(z)}\right)\frac{2\cos(e^z)}{\sin(z)} - \left(\frac{z^2}{\sin^2(z)}\right) \frac{-e^z \sin(e^z) \sin^2(z)}{\sin^2(z)} - \left(\frac{z^2}{\sin^2(z)}\right) \frac{2\cos(e^z) \cos(z) }{\sin(z)}$$

Now when taking the limit, by applying the key fact, all the expressions in parentheses become factors of 1.

$$\lim\limits_{z\to 0}\left[ \frac{2\cos(e^z)}{\sin(z)} - e^z \sin(e^z) - \frac{2\cos(e^z) \cos(z)}{\sin(z)}\right]$$

Plugging in $z=0$ where it doesn't create zeros and simplifying gives $$\lim\limits_{z\to 0}\left[ \frac{2\cos(1)}{\sin(z)} - \sin(1) - \frac{2\cos(1)}{\sin(z)}\right] = -\sin(1)$$

You can follow a very similar process for the other two residues at $z = \pm \pi$, as $ \lim\limits_{z\to c}\left[\,\frac{\sin(z-c)}{z-c}\,\right] = 1 $

Answer 2

The integrand $f(z)=\frac{\cos(e^z)}{\sin^2z}$ has poles of order two at $0$ and $\pm\pi$ lying inside circe $|z|=4$. Computing residues as follows:
$Res[f,z=0]=\lim_{z\rightarrow 0}[z^2 f(z)]'=[\cos (e^z)]'=-\sin (1)$
and $Res[f,z=\pi]=\lim_{z\rightarrow \pi}[(z-\pi)^2 f(z)]'$
$=\lim_{z\rightarrow \pi}\left[(z-\pi)^2\frac{\cos(e^z)}{\sin^2(\pi+(z-\pi))}\right ]'$
$=\lim_{z\rightarrow \pi}\left [(z-\pi)^2\frac{\cos(e^z)}{\sin^2(z-\pi)}\right ]'$
$=\lim_{z\rightarrow \pi}[\cos (e^z)]'$
$=-e^\pi\sin (e^\pi)$
Similarly compute $Res[f,z=-\pi]=-e^{-\pi}\sin(e^{-\pi})$ and then apply Residue theorem.