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.
Best Answer
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!