# Laurent series of $\frac1{\sin^2z}$ around 0

I tried to expand $\frac{z^2}{\sin(z)^{2}}$ using Taylor expansion, but the coefficient involved some limit of $\frac{0}{0}$ and was really difficult to calculate. (I tried to convince myself the constant term of the Taylor expansion is 1 because $\lim_{x\rightarrow 0}(\frac{z^2}{\sin(z)^{2}})$=1 and the coefficients of z with odd degree are $0$ because $\frac{z^2}{\sin(z)^{2}}$ is a even function). But how do I solve those coefficients explicitly...?

Thank you.

You may just write, as $z \to 0$: \begin{align}\frac{1}{\sin^2z}&=\frac{1}{\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}+\mathcal{O}(z^7)\right)^2}\\\\&=\frac{1}{z^2\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\mathcal{O}(z^6)\right)^2} \\\\&=\frac{1}{z^2}\left( 1-2\times \left(-\frac{z^2}{3!}+\frac{z^4}{5!}\right)+3\times \left(-\frac{z^2}{3!}\right)^2+\mathcal{O}(z^6)\right)\\\\ &=\frac{1}{z^2}\left(1+\frac{z^2}{3}+\frac{z^4}{15}+\mathcal{O}(z^6)\right)\\\\ &=\frac{1}{z^2}+\frac{1}{3}+\frac{z^2}{15}+\mathcal{O}(z^4) \end{align} and this gives the first terms of the Laurent series expansion of $\dfrac{1}{\sin^2z}$ near $z=0$.
To obtain the coefficient of order $n$, one may use the Bernoulli numbers, observing that $$\frac{1}{\sin^2z}=-2i\:\partial_z \left(\frac{1}{e^{2iz}-1}\right)$$ then get
$$\frac{1}{\sin^2z}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1} 2^{2n}(2n-1) }{(2n)!}B_{2n}z^{2n-2},\quad 0<|z|<\pi.$$
• @Cure This comes from the Taylor series of $\dfrac{1}{(1+u)^2}$ as $u \to0$ which is equal to $1-2u+3u^2+O(u^3)$, then one puts $u=\frac{z^2}{3}+\frac{z^4}{15}+\mathcal{O}(z^6)$ as $z \to 0$. – Olivier Oloa Nov 11 '17 at 23:06