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**wik_chick88** use the Cauchy residue theorem show that for any integer $\displaystyle m$,

$\displaystyle \oint_C z^{m-1} dz = 2 \pi i \delta_{m,0}$

where $\displaystyle C$ is a circle of radius unity centred at $\displaystyle z = 0$

this part was easy enough.

$\displaystyle \oint_C z^{m-1} dz$

$\displaystyle = \oint_C \frac{z^{m}}{z^1} dz$

$\displaystyle = \frac{2 \pi i}{0!} f^{(0)} (0)$ (by Cauchy residue theorem)

$\displaystyle = 2 \pi i z^m (0)$

$\displaystyle = 0$ if $\displaystyle m = \mathbb{R}$ not 0, or $\displaystyle 2 \pi i$ if $\displaystyle m = 0$

$\displaystyle = 2 \pi i \delta_{m,0}$

now i have to use this result to deduce that, if you are given

$\displaystyle T(z) = \sum^{\infty}_{n=-\infty} z^{-n-2} L_n$

then

$\displaystyle L_n = \frac{1}{2 \pi i} \oint_C z^{n+1} T(z) dz$

and

$\displaystyle (n+2)L_n = -\frac{1}{2 \pi i} \oint_C z^{n+2} T'(z) dz$

where $\displaystyle T'(z)$ is the derivative of $\displaystyle T(z)$ 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 $\displaystyle m\in\mathbb{N}$ and note then that $\displaystyle \displaystyle z^{m+1}T(z)=\sum_{n=-\infty}^{\infty}z^{m+1}z^{-(n+1)}L_n$. Thus, recalling that power series are uniformly convergent, so you can exchange integral and series you get

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