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Math Help - cauchy residue theorem question

  1. #1
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    cauchy residue theorem question

    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??
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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 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}
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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 (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 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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