Compute $\oint \Bigl[ z e^{3/z} + \frac{\cos z}{z^2 (z - \pi )^3} \Bigr] \, dz$

Question

Compute $$\oint \left[ z e^{3/z} + \frac{\cos z}{z^2 (z - \pi )^3} \right] \, dz$$ $$|z| = 5$$

My question is how to do residue at $$\oint ze^{3/z} \, dz $$

Answer 1

Hint:

As @JimmyK4542 suggested, let's expand $e^{3/z}$ using a Laurent series: $$ e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!} $$ We can plug in $3/z$ and get: $$ e^{3/z} = \sum_{n=0}^{\infty}\frac{(\frac{3}{z})^n}{n!} = \sum_{n=0}^{\infty}\frac{3^n}{z^nn!} = 1 + \frac{3}{z} + \frac{9}{2z^2} + \dots $$ From here we get: $$ ze^{3/z} = z + 3 + \frac{9}{2z} + \dots $$

What is the residue of $ze^{3/z}$ then?

Answer 2

You have two poles inside your contour. Namely $z=0$ which is an essential singularity and $z=\pi$ which is a pole of order $3$. For the first pole $z=0$ we will have the Laurent expansion around the point $z=0$

$$ z e^{3/z} + \frac{\cos z}{z^2 (z - \pi )^3} = z\left(1+\frac{3}{z}+ \frac{9}{2!\,z^2}+\dots \right) + \left(\frac{1}{2\pi^3}-\frac{6}{\pi^5} - {\frac {3}{{\pi }^{4}z}}-{\frac {1}{{z}^{2}{\pi }^{3}}}+\dots \right). $$

I think you should be able to find the residue from the above Laurent series which corresponds to the pole $z=0$.

I let you finish the problem. See related problems and techniques.