Toronto Math Forum

Set $ w=z-\pi $

So that $ z=w+\pi $

Using the identity $\sin (w+\pi )=-\sin w$

$f(w+\pi )=\frac{sin (w+\pi )}{((w+\pi )-\pi )^{2}}=\frac{-\sin
w}{w^{2}}$

$-\sin w=\sum_{n=0}^{\infty }{(-1)^{n+1} \frac{w^{2n+1}}{(2n+1)!}}$

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{w^{2n+1-2}}{(2n+1)!}}$

$\frac{-\sin w}{w^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{w^{2n-1}}{(2n+1)!}}$

$
-\frac{1}{w}+\frac{w}{3!}-\frac{w^{3}}{5!}+\frac{w^{5}}{7!}-\frac{w^{7}}{9!}+\ldots
$

Substitute back $w=z-\pi $

$\frac{\sin z}{(z-\pi )^{2}}=\sum_{n=0}^{\infty }{(-1)^{n+1}
\frac{(z-\pi )^{2n-1}}{(2n+1)!}}$

$=-\frac{1}{(z-\pi )}+\frac{(z-\pi )}{3!}-\frac{(z-\pi
)^{3}}{5!}+\frac{(z-\pi )^{5}}{7!}-\frac{(z-\pi )^{7}}{9!}+\ldots $

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

Which is the coefficient of $-\frac{1}{(z-\pi )}$

Or you can use $Res(f,z_{0})=\lim _{z\to z_{0}}(z-z_{0}) f(z)$

$\lim _{z\to \pi}(z-\pi ) \frac{\sin z}{(z-\pi )^{2}}=-1$