Computing the residues of $\frac{\tan z}{z^2-\pi\frac{z}{4}}$

Question

Compute the residues of $$\frac{\tan z}{z^2-\pi\frac{z}{4}}$$

I thought of the following singular points:

$z_k=\frac{\pi}{2}+k\pi\:\:k\in\mathbb{Z}$ which is a pole of order I cannot determine.

$z=\frac{\pi}{4}$ which is a pole of order 1.

I wonder if $z=\infty$ is a singular point however I am not seeing how to compute $\lim_{z\to\infty}\frac{\tan z}{z^2-\pi\frac{z}{4}}$.

However in the book solution the answer was:

$\operatorname{res}_{z=\frac{\pi}{4}}\frac{\tan z}{z^2-\pi\frac{z}{4}}=\frac{4}{\pi}$

Questions:

Why is the solution ruling out $z_k=\frac{\pi}{2}+k\pi\:\:k\in\mathbb{Z}$? Is $z=\infty$ a singular point?

Thanks in advance!

Answer 1

First of all, the limit $\lim_{z\to\infty}\dfrac{\tan z}{z^2-z\frac\pi4}$ doesn't exist.

And indeed that function has a simple pole at every point of the form $\dfrac\pi2+k\pi$. So, the answer from your textbook is wrong.

Answer 2

You can write $f(z)=g(z)/(z-π/4)$ where $g(z)=tan z / z $. As $g(π/ 4)$ not zero 0 and analytic there, $π/4$ is pole of order 1. So residue is simply $ g(π /4)=4/ π$