How can I solve this integral with complex numbers

Question

$$\int_{|z-1| = 1/4}\frac{1}{(z-1)\cos(\pi z)}dz$$

$$\int_{|z-1| = 3/4}\frac{1}{(z-1)\cos(\pi z)}dz$$

$z$ here is a complex number. How can I solve this integral by using Residues.

Answer 1

HINT:

The integrand has simple poles at $z=1$, $z=\pm (2n-1)/2$ for integer $n$. The residues are given by

$$\lim_{z\to z_p}\frac{(z-z_p)}{(z-1)\cos(\pi z)}$$

where $z_p$ is a pole. Use L'Hospital's Rule to evaluate the residues.

Now, which poles, if any, are enclosed when $|z-1|< 1/4$ and which poles are enclose when $|z-1|< 3/4$?