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Math Help - Mittag-Leffler Cotangent expansion proof Complex analysis

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    Mittag-Leffler Cotangent expansion proof Complex analysis

    Hello! I have this problem for my assignment, and I'm stuck at step 2, any help would be greatly appreciated thanks!


    Prove the formula claimed in the
    lecture as follows:
    Mittag-Leffler Cotangent expansion proof Complex analysis-untitled.jpg

    1)We know, looking at residues, that the difference:
    Mittag-Leffler Cotangent expansion proof Complex analysis-untitled-copy.jpg
    is an entire function (that is, it is holomorphic everywhere on C. Compute the derivative of this difference (using cot = -1/sin^2) by reordering the sum.

    2)Prove that for |Im(z)|-> inf, d'(z) -> 0. By the previous step you can assume w.l.o.g. that 0 < Re(z) <= 1! (Consider
    the term involving sin and the infinite sum separately.)


    I can solve the rest of the problem but I can't seem to prove that the lim as the imaginary part of zgoes to infinity makes d'(z) -> 0

    The derivative I get is:
    d'(z)= -pi^2/ sin^2(pi*z) +1/z^2+ pi/z *coth(z*pi)

    Thank you!!
    Last edited by Jimena; November 25th 2010 at 06:35 PM.
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  2. #2
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    Quote Originally Posted by Jimena View Post
    Hello! I have this problem for my assignment, and I'm stuck at step 2, any help would be greatly appreciated thanks!


    Prove the formula claimed in the
    lecture as follows:
    Click image for larger version. 

Name:	Untitled.jpg 
Views:	31 
Size:	15.4 KB 
ID:	19857

    1)We know, looking at residues, that the difference:
    Click image for larger version. 

Name:	Untitled - Copy.jpg 
Views:	35 
Size:	16.3 KB 
ID:	19858
    is an entire function (that is, it is holomorphic everywhere on C. Compute the derivative of this difference (using cot = -1/sin^2) by reordering the sum.

    2)Prove that for |Im(z)|-> inf, d'(z) -> 0. By the previous step you can assume w.l.o.g. that 0 < Re(z) <= 1! (Consider
    the term involving sin and the infinite sum separately.)


    I can solve the rest of the problem but I can't seem to prove that the lim as the imaginary part of zgoes to infinity makes d'(z) -> 0

    The derivative I get is:
    d'(z)= -pi^2/ sin^2(pi*z) +1/z^2+ pi/z *coth(z*pi)
    I think you would be better off leaving the derivative in the form \displaystyle d'(z) = -\frac{\pi^2}{\sin^2\pi z} + \frac1{z^2} + \sum_{k\in\mathbb{Z}\setminus\{0\}}\frac1{(z-k)^2}.

    In the sum, group the terms with index k and ľk (that's what the hint about reordering the sum is getting at!), to make it \displaystyle \sum_{k=1}^\infty \frac{2(z^2+k^2)}{(z^2-k^2)^2}. In that form, it's easier to see what happens as |Im(z)| goes to infinity.

    This is all spelt out in Ahlfors' Complex Analysis (Section 5.2 in the edition that I have). If you want to get to grips with complex analysis, Ahlfors is really an essential possession.
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