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Thread: cauchy residue theorem question

  1. #1
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    cauchy residue theorem question

    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??
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: cauchy residue theorem question

    Quote Originally Posted by wik_chick88 View Post
    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


    $\displaystyle \displaystyle \begin{aligned}\oint_C z^{m+1}T(z)\; dz &= \oint_C \sum_{n=-\infty}^{\infty}z^{m+1}z^{-(n+1)-1}L_n\\ &= \sum_{n=-\infty}^{\infty}\oint_C z^{m-(n+1)}L_n\\ &= \sum_{n=-\infty}^{\infty}2\pi i \delta_{n,m}L_n\\ &= 2\pi i L_m\end{aligned}$
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  3. #3
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    Re: cauchy residue theorem question

    thankyou so much! with your help it was easy to do the second bit as well the $\displaystyle (n+2)L_n$ part so thankyou thankyou thankyou!
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  4. #4
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    Re: cauchy residue theorem question

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

    any ideas????
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