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

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

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

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.

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

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

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

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

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!