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Thread: singularities and residues

  1. #1
    Nov 2012

    singularities and residues

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

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

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

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

    A lats question: is it correct to say: the poles $\displaystyle z_k$ accumulates to $\displaystyle \infty$, hence $\displaystyle \infty$ is not an isolated singularity, thus i cannot compute $\displaystyle Res(f;\infty)$?
    Last edited by tenderline; Nov 27th 2012 at 11:58 AM.
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