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Homework Help: A: Reciprocal series, B: Laurent Series and Cauchy's Formula

  1. Jan 13, 2016 #1
    Problem A now solved!

    Problem B:
    I am working with two equations:

    The first gives me the coefficients for the Laurent Series expansion of a complex function, which is:

    [itex] f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n [/itex]

    This first equation for the coefficients is:

    [itex] a_n = \frac{1}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz [/itex]

    The second equation is Cauchy's Integral formula for the nth derivative of a complex function:

    [itex] f^{(n)}(z_0) = \frac{n!}{2πi} \oint \frac{f(z)}{(z-z_0)^{n+1}} \, dz [/itex]

    My problem is as follows:

    We can clearly re-arrange these two equations to get the following expression:

    [itex] a_n = \frac{f^{(n)}(z_0)}{n!} [/itex]

    Now, in this instance, [itex] f(z) = \frac{z^3 + 2z^2 + 4}{(z-1)^3} [/itex].

    If we try to calculate [itex] a_n [/itex] for this function, it keeps getting sent to zero, due to the function being evaluated at [itex] z_0 [/itex]!

    What am I missing? Where have I gone wrong in this derivation/problem?

    Ultimately in this problem I'm trying to find the Laurent Series expansion of the function.

    Any help/tips would be much appreciated!

    Last edited: Jan 13, 2016
  2. jcsd
  3. Jan 13, 2016 #2


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    2017 Award

    Staff: Mentor

    You first take the derivative, then plug in a fixed z0. In general, the result should not be zero.
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