$$\int_{|z-1| = 1/4}\frac{1}{(z-1)\cos(\pi z)}dz$$
$$\int_{|z-1| = 3/4}\frac{1}{(z-1)\cos(\pi z)}dz$$
$z$ here is a complex number. How can I solve this integral by using Residues.
How can I solve this integral with complex numbers
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$\begingroup$
calculus
complex-analysis
complex-numbers
residue-calculus
complex-integration
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1You may mean $\cos(\pi z)$, right? Just simply check where the zeros of $z-1$ and $\cos(\pi z)$ are and their order.. – 2017-01-27
1 Answers
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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 $|z-1|< 3/4$?