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.
 A: 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
$$
\frac1{(z-k)^2}.
$$
A: We can try the following using the well known power series for sine:
$$\sin\pi z=(-1)^k\sin\left[\pi(z-k)\right]\implies$$
$$ \sin^2\pi z=\sin^2\pi(z-k)=\left(\sum_{n=0}^\infty(-1)^n\frac{\pi^{2n+1}(z-k)^{2n+1}}{(2n+1)!}\right)^2=$$
$$=\left(\pi(z-k)-\frac{\pi^3(z-k)^3}{3!}+\ldots\right)\left(\pi(z-k)-\frac{\pi^3(z-k)^3}{3!}+\ldots\right)=$$
$$=\pi^2(z-k)^2-\frac{\pi^4}3(z-k)^4+\mathcal O((z-k)^6)\implies$$
$$\frac{\pi^2}{\sin^2\pi z}=\frac{\pi^2}{\pi^2(z-k)^2\left(1-\frac{\pi^2}3(z-k)^2+\mathcal O((z-k)^3)\right)}=$$
$$=\frac{\pi^2}{\pi^2(z-k)^2}\left(1+\frac{\pi^2(z-k)^2}{3}+\ldots\right)$$
and the principal part is thus $\;\cfrac1{(z-k)^2}\;$ , which was expectable as we have double poles here.
