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Math Help - Laurent Series Problem

  1. #1
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    Laurent Series Problem

    Hi

    Can someone tell me what are the steps in finding the Laurent series expansion of f(z) = \frac{1}{z(z-1)^2} about z=0 and z=1

    i found the binomial expansion when z=0 is \frac{1}{z}-2-3z-4z^2

    P.S
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Laurent Series Problem

    Quote Originally Posted by Paymemoney View Post
    i found the binomial expansion when z=0 is \frac{1}{z}-2-3z-4z^2
    That is not correct. Consider \frac{1}{1-z}=\sum_{n=0}^{+\infty}z^n and take derivatives in both sides. Around z=1 , use the substitution u=z-1.
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  3. #3
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    Re: Laurent Series Problem

    after doing the problem again i got \frac{1}{z}-2+3z-4z^2, however answers are \frac{-1}{z}-2-3z-4z^2
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