cauchy residue theorem question

**wik_chick88** · October 10th 2011, 04:11 PM

use the Cauchy residue theorem show that for any integer $m$ ,
$\oint_C z^{m-1} dz = 2 \pi i \delta_{m,0}$
where $C$ is a circle of radius unity centred at $z = 0$

this part was easy enough.
$\oint_C z^{m-1} dz$
$= \oint_C \frac{z^{m}}{z^1} dz$
$= \frac{2 \pi i}{0!} f^{(0)} (0)$ (by Cauchy residue theorem)
$= 2 \pi i z^m (0)$
$= 0$ if $m = \mathbb{R}$ not 0, or $2 \pi i$ if $m = 0$
$= 2 \pi i \delta_{m,0}$

now i have to use this result to deduce that, if you are given
$T(z) = \sum^{\infty}_{n=-\infty} z^{-n-2} L_n$
then
$L_n = \frac{1}{2 \pi i} \oint_C z^{n+1} T(z) dz$
and
$(n+2)L_n = -\frac{1}{2 \pi i} \oint_C z^{n+2} T'(z) dz$
where $T'(z)$ is the derivative of $T(z)$ with respect to z

i have no idea how to even start the second part of this question!! any tips??

**Drexel28** · October 10th 2011, 04:37 PM

Originally Posted by wik_chick88

use the Cauchy residue theorem show that for any integer $m$ ,
$\oint_C z^{m-1} dz = 2 \pi i \delta_{m,0}$
where $C$ is a circle of radius unity centred at $z = 0$

this part was easy enough.
$\oint_C z^{m-1} dz$
$= \oint_C \frac{z^{m}}{z^1} dz$
$= \frac{2 \pi i}{0!} f^{(0)} (0)$ (by Cauchy residue theorem)
$= 2 \pi i z^m (0)$
$= 0$ if $m = \mathbb{R}$ not 0, or $2 \pi i$ if $m = 0$
$= 2 \pi i \delta_{m,0}$

now i have to use this result to deduce that, if you are given
$T(z) = \sum^{\infty}_{n=-\infty} z^{-n-2} L_n$
then
$L_n = \frac{1}{2 \pi i} \oint_C z^{n+1} T(z) dz$
and
$(n+2)L_n = -\frac{1}{2 \pi i} \oint_C z^{n+2} T'(z) dz$
where $T'(z)$ is the derivative of $T(z)$ with respect to z

i have no idea how to even start the second part of this question!! any tips??

I would say, fix an $m\in\mathbb{N}$ and note then that $\displaystyle z^{m+1}T(z)=\sum_{n=-\infty}^{\infty}z^{m+1}z^{-(n+1)}L_n$ . Thus, recalling that power series are uniformly convergent, so you can exchange integral and series you get

$\displaystyle \begin{aligned}\oint_C z^{m+1}T(z)\; dz &= \oint_C \sum_{n=-\infty}^{\infty}z^{m+1}z^{-(n+1)-1}L_n\\ &= \sum_{n=-\infty}^{\infty}\oint_C z^{m-(n+1)}L_n\\ &= \sum_{n=-\infty}^{\infty}2\pi i \delta_{n,m}L_n\\ &= 2\pi i L_m\end{aligned}$

**wik_chick88** · October 10th 2011, 04:59 PM

thankyou so much! with your help it was easy to do the second bit as well the $(n+2)L_n$ part so thankyou thankyou thankyou!

**wik_chick88** · October 10th 2011, 05:10 PM

this is a different question but is the same kind of stuff...
you are given $L_n L_m - L_m L_n = \frac{1}{(2 \pi i)^2} \oint_{C_0} dw \ w^{m+1} \oint_{C_w} dz \ z^{n+1}T(z)T(w)$
where $C_w$ is a circle of radius unity centred at $z = w$ and $C_0$ is a circle of radius unity centred at $w = 0$ and
$T(z)T(w) = \frac{c}{2(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{T'(w)}{z-w}$
with $c$ being a constant and $T'(w)$ being the derivative of $T(w)$ with respect to $w$
show that $L_n L_m - L_m L_n = (n-m)L_{n+m} + \frac{1}{12}cn(n^2 - 1)\delta_{n+m,0}$

any ideas????

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