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Math Help - isolated point, Cauchy integral formula

  1. #1
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    isolated point, Cauchy integral formula

    Let z_0 be an isolated point of a function f and suppose that f(z)=\frac{\phi(z)}{(z-z_0)^m}, where m is a positive integer and \phi(z) is analytic and nonzero at z_0. By applying the extended form of the Cauchy integral formula to the function \phi(z), show that \text{Res}_{z=z_0}=\frac{\phi^{(m-1)}(z_0)}{(m-1)!}.

    I do not see how to do this. Our book says:
    Since there is a neighborhood |z-z_0|< \epsilon throughout which \phi(z) is analytic, the contour used in the extended Cauchy integral formula can be the positively oriented circle |z-z_0|< \frac{\epsilon}{2}. How do I use this suggestion? I don't see that now. Thanks in advance.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The defintion of 'residue' of an f(z) which has a pole of order m in z=z_{0} and is analytic elsewhere in a region around z_{0} is ...

    Res_{z=z_{0}} f(z) = \lim_{z \rightarrow z_{0}} \frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}} \{(z-z_{0})^{m}\cdot f(z)\} (1)

    Now if is...

    f(z)= \frac{\phi(z)}{(z-z_{0})^{m}} (2)

    ... where \phi(*) is analytic in z=z_{0} the (1) gives to You...

    Res_{z=z_{0}} f(z) = \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}} \phi(z)_{z=z_{0}} (3)

    Kind regards

    \chi \sigma
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