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.

  • $\begingroup$ $f$ has period $1$, so the principal part at $k$ is the translation of the principal part at $0$. As $\dfrac{\sin (\pi z)}{\pi z} = 1 - O(z^2)$, we have $\dfrac{\pi z}{\sin (\pi z)} = 1 + O(z^2)$, whence $\dfrac{\pi^2z^2}{(\sin (\pi z))^2} = 1 + O(z^2)$ and hence $\dfrac{\pi^2}{(\sin (\pi z))^2} = \dfrac{1}{z^2} + O(1)$. $\endgroup$ – Daniel Fischer Jul 2 '16 at 12:21

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 $$ \frac1{(z-k)^2}. $$

  • $\begingroup$ How do you know that the bottom portion converges uniformly? $\endgroup$ – adam Jul 1 '16 at 22:18
  • $\begingroup$ @adam Please see my edit, I've used that, as $u \to 0$, $\frac1{1-u}=1+u+O(u^2)$ giving $\frac1{1-\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4)}=1+\frac{\pi^2 \varepsilon^2}{3}+O(\varepsilon^4)$. Thanks. $\endgroup$ – Olivier Oloa Jul 1 '16 at 22:23
  • $\begingroup$ @Oliver Oloa I understand that but we need to know we have uniform convergence before we do that right? $\endgroup$ – adam Jul 1 '16 at 22:24
  • $\begingroup$ Yes, to prove it once for all. $\endgroup$ – Olivier Oloa Jul 1 '16 at 22:26

We can try the following using the well known power series for sine:

$$\sin\pi z=(-1)^k\sin\left[\pi(z-k)\right]\implies$$

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


$$=\pi^2(z-k)^2-\frac{\pi^4}3(z-k)^4+\mathcal O((z-k)^6)\implies$$

$$\frac{\pi^2}{\sin^2\pi z}=\frac{\pi^2}{\pi^2(z-k)^2\left(1-\frac{\pi^2}3(z-k)^2+\mathcal O((z-k)^3)\right)}=$$


and the principal part is thus $\;\cfrac1{(z-k)^2}\;$ , which was expectable as we have double poles here.

  • 1
    $\begingroup$ I like this one! $\endgroup$ – adam Jul 1 '16 at 22:27

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