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Thread: singularities and residues

  1. #1
    Nov 2012

    singularities and residues

    Consider the function f(z)=\frac{z^3}{1-cosh(z)}. Find its singularities and compute residues.

    I know the denominator vanishes for z_k=2k\pi i, k integer. I first consider k=0, so the function is analytic in $ 0<|z|<2\pi, and i can write in this punctured disc the following Laurent expansion:

    starting from cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!} i get cosh(z)=cos(iz)=\sum_{n=0}^{\infty}\frac{z^{2n}}{(  2n)!}Hence 1-cosh(z)=\frac{z^2}{2!}-\frac{z^4}{4!}\ldots
    thus i can write \frac{1}{1-\cosh(z)}=\frac{1}{-\frac{z^2}{2!}-\ldots}=-\frac{2}{z^2(1-h)}=-\frac{2}{z^2}(1+h+h^2\ldots) where h=-\frac{2z^2}{2!}-\frac{2z^4}{4!}-\ldots.
    So we have \frac{1}{1-cosh(z)}=-\frac{2}{z^2}+\frac{4}{4!}+ higther terms.
    Finally, we get \frac{z^3}{1-cosh(z)}=-2z+\frac{4z^3}{4!}+ higther terms, fromw which i desume that z_0=0 is a removable singularity for f.

    But now i don't knoe hoe to deal with z_k with k\neq 0. I imagine those to be all poles of order 2 for f, but how to prove?

    A lats question: is it correct to say: the poles z_k accumulates to \infty, hence \infty is not an isolated singularity, thus i cannot compute Res(f;\infty)?
    Last edited by tenderline; Nov 27th 2012 at 12:58 PM.
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