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Complex functions and poles

  1. May 14, 2015 #1
    Suppose we have a complex function [tex] f(z)[/tex] with simple poles on the complex plane, and we know exactly where these poles are located (but we don't know how the function depends on z) Is there any way to build up the exact form of [tex] f(z)[/tex] just from its poles?
     
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  3. May 14, 2015 #2

    Svein

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    No. You can conclude that the function is of the form [itex] H(x)+\sum_{n}\frac{c_{n}}{(z-z_{n})^{k_{n}}}[/itex], where H is analytic, zn are the poles and the cn and kn are constants (the kn are positive integers).
     
  4. May 14, 2015 #3

    mfb

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    For every function f(z), 2*f(z), z*f(z) and so on have the same poles (the last one assuming 0 is not a pole).
     
  5. May 14, 2015 #4
    Yes, thanks for both replies, I think the first one is related with Mittag Leffers theorem, the poles I am talking about follow a law, for example, a harmonic series on the real axis. But the sum over all this poles gives a complicated function the complex plane. To be clear, I am trying to evaluate a function of the type: [tex] f(z)=\sum_{n=-\infty}^{+\infty}\frac{1}{z-z_{n}} [/tex] where the zn poles can follow a law, for example [tex] z_{n}=A*(n+\frac{1}{2})^{2} [/tex] where A is constant. The problem is that the series depending on these poles can give as a result really complicated functions of z, I just wanted to know if there is an analytical way to carry out these sums.
     
  6. May 14, 2015 #5

    mfb

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    Some poles will match (e. g. n=1, n=-2).

    A sum like$$\sum_{n=1}^\infty \frac{1}{1+(n+c)^2}$$ should have an analytic result, and you can modify your sum to have that shape.
     
  7. May 14, 2015 #6
    Thats actually what I have done, and yes it has to have an analytical form, I was just wondering about an analytical method to arrive at it, rather than using Mathematica, just to know if there exist a general method or something, many thanks for the replies
     
  8. May 14, 2015 #7

    Svein

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    Picked up my trusty Ahlfors (Complex analysis) and he shows that [itex]\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^{2}}=\frac{\pi^{2}}{\sin^{2}\pi z} [/itex] and [itex]\frac{1}{z}+\sum_{n\neq 0}(\frac{1}{z-n}+\frac{1}{n})=\pi\cot(\pi z) [/itex]. Might give you some ideas.
     
  9. May 14, 2015 #8

    mfb

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  10. May 24, 2015 #9
    Many thanks for the replies!
     
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