Residue of $\frac{\cos(\frac{\pi}{z-1})}{z^2 \sin z}$ at $z=1$

Question

Residue of $$\frac{1}{z^2 \sin z}\cos\left(\frac{\pi}{z-1}\right)$$ at $z=1$.

More importantly, I don't even know whether it exists or not. The one who creates this question has made questions that are unsolvable.

I have tried some methods while they are not so successful.

1. Wolfram alpha. It doesn't even give an answer this times.

2. Series expansion. But this turns out to be too ugly. Expanding $\cos$, $z^2$ and $\sin $ respectively, and evaluate the coefficient of $\frac{1}{z-1}$ seems impossible and silly (without aid of matlab).

3. see if it is a removable singularity. Considering $\lim_{z \to 1} (z-1) f(z)$, I once thought I made it by $-1 \leq \cos z \leq 1$, but this inequality doesn't apply in complex.

Please help.

Answer 1

Instead of computing the residue in the given point, compute the residues in the other singularities. Given $f(z)=\frac{\cos\frac{\pi}{z-1}}{z^2\sin z},$ we have:

$$ \operatorname{Res}_{z=k\pi} f(z) = \frac{(-1)^k}{\pi^2 k^2} \cos\frac{\pi}{\pi k -1} $$ for any $k\in\mathbb{Z}\setminus 0$ and:

$$ \operatorname{Res}_{z=0} f(z) = \frac{3\pi^2-1}{6},$$ so the residue you want to compute is given by a convergent series: $$ \operatorname{Res}_{z=1} f(z) = \frac{1-3\pi^2}{6}-\sum_{k=1}^{+\infty}\frac{(-1)^k}{\pi^2 k^2}\left(\cos\frac{\pi}{\pi k-1}+\cos\frac{\pi}{\pi k+1}\right) = \color{red}{-4.7143885\ldots}.$$

Thanks to Daniel Fischer and Ron Gordon, this holds because the sum of all the residues is zero, since for any positive number $n$, $$ \oint_{|z|=\frac{\pi}{2}+n\pi}f(z)\,dz = O\left(\frac{1}{n}\right).$$ Moreover, since for any positive natural number $k$: $$ \operatorname{Res}_{z=k\pi}f(z)=\frac{1}{2\pi i}\left(\int_{|z|=(k+1/2)\pi}f(z)\,dz-\int_{|z|=(k-1/2)\pi}f(z)\,dz\right) = O\left(\frac{1}{k^2}\right), $$ we have: $$ \oint_{|z|=R}f(z)\,dz=O\left(\frac{1}{R}\right) $$ for any $R\in\mathbb{R}_{>1}\setminus\pi\mathbb{N}.$