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Complex Analysis: Is my proof of an easy theorem correct?

  1. Nov 13, 2013 #1
    Theorem: If a function f(z) has a zero of nth order at z0, then the function h(z)/f(z) has a pole of order n at z0 (where h(z) is analytic at ##z_0##).

    Can somebody explain this theorem for me? It isn't proved in my book because it's so "easy", but I don't get it? Is the sketch of the proof something like this?

    f(z) has a taylor expansion around z0 which begins on the nth term (since ##f^{(1)}(z_0),f^{(2)}(z_0),....,f^{(n-1)}(z_0) = 0##, and thus when you find the taylor expansion of f(z), you divide 1 by said taylor series to gain the expansion of 1/f(z). After splitting up the expansion of 1/f(z) into several different fractions (with the help of some fancy algebra), you will gain a laurent series of order n.

    But how do I get to the last part? I am completely confused right now..
     
    Last edited: Nov 13, 2013
  2. jcsd
  3. Nov 13, 2013 #2

    Dick

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    You should also specify that h(z) is nonzero at z0. If f(z) has a zero of nth order at z0, then you can write it as f(z)=(z-z0)^n*g(z) where g(z) is analytic and nonzero at z0. Does that make it clearer?
     
  4. Nov 13, 2013 #3
    Damn, of course!! Thanks now I understand the theorem.

    Since g(z) is analytic and nonzero, it can be expanded as a taylor series, and thus the largest order negative polynomial possible in the total series is n!!
     
    Last edited: Nov 13, 2013
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