TITLE: Computing the residues of $\dfrac{1}{\sin^2 z}$.
QUESTION [6 upvotes]: I'm having a hard time correctly computing the residues of $\dfrac{1}{\sin^2 z}$. I 
know the poles occur at $k\pi$, with order $2$. 
By a Taylor expansion I can rewrite $\sin z=\cos k\pi(z-k\pi)+f_2(z)(z-k\pi)^2$, and so 
$$
\sin^2 z=(z-k\pi)^2(\cos k\pi+f_2(z)(z-k\pi))^2.
$$
I want to calculate the residue with Cauchy's Integral Theorem, so 
$$
\text{Res}(f,k\pi)=\frac{1}{2\pi i}\int_{|z-k\pi|=1}\frac{dz}{(z-k\pi)^2[\cos k\pi
+f_2(z)(z-k\pi)]^2}.
$$
This should equal the derivative of $(\cos k\pi+f_2(z)(z-k\pi))^{-2}$ evaluated at $k

\pi$. The derivative comes out to be
$$
-2(\cos k\pi+f_2(z)(z-k\pi))^{-3}(f'_2(z)(z-k\pi)+f_2(z))
$$ 
and evaluates to $\dfrac{-2f_2(k\pi)}{(\cos k\pi)^3}$. Apparently the residue should 
just be $0$, but I don't see how to conclude this. What am I missing to know $f_2(k\pi)=0$?

REPLY [11 votes]: $\sin(z)=z-z^3/6+\dots$ so
$$
\begin{align}
\frac{1}{\sin^2(z)}
&=\frac{1}{z^2}\left(1-z^2/6+\dots\right)^{-2}\\
&=\frac{1}{z^2}\left(1+z^2/3+\dots\right)\\
&=\frac{1}{z^2}+\frac13+\dots
\end{align}
$$
Thus, the residue at $z=0$ is $0$ since there is no $\dfrac{1}{z}$ term. The others follow by periodicity.