Finding the principal part for the Laurent series How can I find the principal part of the Laurent series for  $f(z)=\dfrac{\pi^2}{(\sin \pi z)^2}$ centered at $k$ where $k \in \mathbb{Z}$.
I think there are two ways to do it either use the formula $a_k$ for Laurent coefficients or expand manipulate $\sin^2 \pi z$ and solve for the coefficients. I am not sure if any of these ways are efficient.

By setting $z:=k+\varepsilon$, $k \in \mathbb{Z}$, $\varepsilon \to 0$, one has by the Taylor series expansion:
$$
\sin^2 \pi z=\left(\sin (\pi k+\pi\varepsilon)\right)^2=\sin^2(\pi\varepsilon)=\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6)
$$ giving
$$
f(z)=\frac{\pi^2}{\pi^2 \varepsilon^2-\frac{\pi^4 \varepsilon^4}{3}+O(\varepsilon^6)}=\frac1{\varepsilon^2(1-\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4))}=\frac1{\varepsilon^2}+\frac{\pi ^2}{3}+O(\varepsilon^2)
$$ that is, as $z \to k$,
$$
f(z)=\frac{\pi^2}{\sin^2 \pi z}=\frac1{(z-k)^2}+\frac{\pi ^2}{3}+O((z-k)^2),
$$ 
the principal part is
$$
\frac1{(z-k)^2}.
$$
