Find the Laurent series and residue of $\frac{z}{(\sin(z))^2}$ at $z_0 = 0$.

Question

Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

$$\frac{z}{(\sin(z))^2}\quad \text{at}\quad z_0 = 0 \quad\text{four terms of the Laurent series}$$

I am not sure how to approach this question. Can anyone help me with this? Thank you.

Answer 1

The Laurent series for $\sin z$ about $z_0 = 0$ is given by it's Taylor series about $z_0 = 0$ because sine is entire. Then, you have $z$ over the Taylor series of $\sin(z)$. Assuming you want to find the Laurent series to find the residue, I will first do everything approximately removing terms that won't contribute to the residue. i.e.

$\frac{z}{\sin^2(z)} = \frac{z}{\left(z-\frac{z^3}{3!}+\ldots\right)\left(z-\frac{z^3}{3!}+\ldots\right)} = \frac{z}{(z^2 - \frac{z^4}{3}+\ldots)} = \frac{z}{z^2\left(1-\frac{z^2}{3}+\ldots\right)} = \frac{1}{z(1-w)}$

where $1-w = 1-\frac{z^2}{3}+\ldots$. Then $w$ is a small term because we are expanding about $z_0=0$. So we can expand $1-w$,

$\frac{1}{1-w} = 1 + w + w^2 + \ldots$,

hence we find

$\frac{z}{\sin^2(z)} = \frac{1}{z}\left(1+w + w^2 + \ldots\right) = \frac{1}{z}(1+ (\frac{z^2}{3} + \ldots))$,

where everything has been valid if we care only about the residue. Hence, $Res\left(\frac{z}{\sin^2(z)} \right) = 1$.

If instead you actually do require a Laurent series, we can let $w = \frac{1}{z}$ and find the Laurent series of

$\frac{(1/w)}{\sin^2(1/w)}$ so we have

$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)}. $

Note that $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, we use this identity ($m=-n$) to find

$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{1/w}{ \left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)\left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)}$.

Next we multiply the series

$\frac{(1/w)}{ \sum_{n=-\infty}^0 \sum_{m=-\infty}^0 (-1)^{n+m}\frac{w^{2(n+m)+2}}{(2(-n)+1)!(2(-m)+1)!}}$

where I have used that $(-1)^n = (-1)^{-n}$ Now let $l=n+m$, yielding

$\frac{(1/w)}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+2}}{(2(m-l)+1)!(2(-m)+1)!}} = \frac{1}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+1}}{(2(m-l)+1)!(2(-m)+1)!}}$.

And you find the residue when $l=0$, so you find the residue to be 1, again!

Answer 2

Since $$ \frac{z}{\sin^2z}=\frac{1}{z}\biggl(\frac{\sin z}z\biggr)^{-2}, $$ then $$ \frac{z}{\sin^2z}=\frac1z+\frac1z\sum_{m=1}^{\infty}(-1)^m\Biggl[\sum_{k=1}^{2m}\frac{(2)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}\Biggr]\frac{(2z)^{2m}}{(2m)!} $$ for $0<|z|<\pi$, where the rising factorial $(r)_k$ is defined by \begin{equation*}%\label{rising-Factorial} (r)_k=\prod_{\ell=0}^{k-1}(r+\ell) = \begin{cases} r(r+1)\dotsm(r+k-1), & k\ge1\\ 1, & k=0 \end{cases} \end{equation*} and $R\bigl(2m+j,j,-\frac{j}2\bigr)$ denotes weighted Stirling numbers of the second kind which can be explicitly computed by \begin{equation}\label{S(n,k,x)-satisfy-eq} R(n,k,r)=\frac1{k!}\sum_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n \end{equation} for $r\in\mathbb{R}$ and $n\ge k\ge0$.

Consequently, it is clear that the residue is $1$.

The above derivation used the following theorem.

Theorem. For $r\in\mathbb{R}$, assume that the value of the power function $\bigl(\frac{\sin x}{x}\bigr)^{r}$ at the point $x=0$ is $1$.

1. For $r\ge0$, the series expansion \begin{equation}\label{recip-sin-ser-closed-eq}\tag{SEH} \biggl(\frac{\sin x}{x}\biggr)^r=1+\sum_{m=1}^{\infty}(-1)^m\Biggl[\sum_{k=1}^{2m}\frac{(-r)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}\Biggr]\frac{(2x)^{2m}}{(2m)!} \end{equation} is convergent in $x\in\mathbb{R}$.
2. For $r<0$, the series expansion \eqref{recip-sin-ser-closed-eq} is convergent in $x\in(-\pi,\pi)$.

The residue $1$ can also be verified simply by the limit $$ \lim_{z\to0}\biggl(\frac{\sin z}z\biggr)^{-2}=1 $$ and then the relation $$ \frac{z}{\sin^2z}\sim\frac{1}{z}, \quad z\to0. $$ For knowledge of weighted Stirling numbers of the second kind, please read Carlitz's paper [1].

For the proof of the above theorem, please read the arXiv preprint [2] below.

References

1. L. Carlitz, Weighted Stirling numbers of the first and second kind, I, Fibonacci Quart. 18 (1980), no. 2, 147--162.
2. F. Qi, Series expansions for any real powers of (hyperbolic) sine functions in terms of weighted Stirling numbers of the second kind, arXiv (2022), available online at https://arxiv.org/abs/2204.05612 or https://doi.org/10.48550/arXiv.2204.05612.