Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $h=-\frac{2z^2}{4!}-\frac{2z^4}{6!}-\ldots$. 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, from which i desume that $z_0=0$ is a removable singularity for f.

But now i don't know how 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 last question: is it correct to say: the poles $z_k$ accumulates to $\infty$, hence $\infty$ is not an isolated singularity, thus i cannot compute $Res(f;\infty)$?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

You can perform the same procedure. The Taylor series for cosine near $z = 2k\pi$ is the same as its series at $z = 0$ because of periodicity. So

$$ \cos z = \sum_{n=0}^{\infty} (-1)^n \frac{(z-2k\pi)^{2n}}{(2n)!}, $$


$$ \begin{align*} \cosh z &= \sum_{n=0}^{\infty} (-1)^n \frac{(iz-2k\pi)^{2n}}{(2n)!} \\ &= \sum_{n=0}^{\infty} \frac{(z+i2k\pi)^{2n}}{(2n)!}. \end{align*} $$

Then rewrite the numerator of $f(z)$ as

$$ \begin{align*} z^3 &= (z+i2k\pi-i2k\pi)^3 \\ &= (z+i2k\pi)^3 - i6k\pi(z+i2k\pi)^2 - 12 k^2\pi^2 (z + i2k\pi) + i2k^3\pi^3. \end{align*} $$

share|improve this answer
I guess this gives you the Laurent series for $f$ near $z = -i2k\pi$. If you'd like, you can just replace $k$ by $-k$ to get the series near $z = i2k\pi$. –  Antonio Vargas Nov 27 '12 at 22:04
thanks, using your suggestion i computed $res(f,z_k)=24k^2\pi^2$, which i hope to be correct –  Federica Maggioni Nov 28 '12 at 11:46
@FedericaMaggioni yup, you got it :) –  Antonio Vargas Nov 28 '12 at 17:37
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.