I'm trying to find the poles and residues of: $$f(z) = \frac{\cos(z)-1}{(e^z - 1)^2}$$

I can see that this has a removable singularity at $z=0$ and double poles at $z=2k \pi i$. I'm having trouble finding the residues of these. In general, when we have more than a single pole, i.e a double pole, triple pole etc. is it best to just try and find the Laurent expansion about the pole?

In this case, writing $(e^z + 1)^{-2} = (e^{z+2k\pi i - 2k \pi i} - 1)^{-2}$ and then expanding this: $$(e^{2k \pi i}(e^{z- 2k \pi i}) + 1)^{-2} = ((e^{2k \pi i}(1+(z-2k \pi i) + \cdots -1)^{-2}$$

I can't seem to arrive at a sensible answer proceeding like this, and I'm not even sure it is the right method.

Any help is much appreciated


"In general, when we have more than a single pole, i.e a double pole, triple pole etc. is it best to just try and find the Laurent expansion about the pole?"

In my experience, for small orders, usually it's easier to use a well known formula.

Given an holomorphic function $\varphi$ with a pole order $k$ at $z_0$, one has, around $z_0$,

$$\varphi (z)=\sum \limits_{n=-k}^\infty \left(c_n(z-z_0)^n\right),$$ for a certain complex sequence $\left(c_n\right)_{n\in \mathbb Z}$, where $\forall n\in \mathbb Z(n<-k\implies c_n=0)$.

Multiply by $(z-z_0)^k$ to get $$\begin{align}(z-z_0)^k\varphi(z)&=\sum \limits_{n=-k}^\infty \left(c_n(z-z_0)^{n+k}\right)\\ &=c_{-k}+c_{-k+1}(z-z_0)+\ldots +\underbrace{c_{-1}}_{\text{Res}(\varphi , z_0)}(z-z_0)^{-1+k}+\ldots .\end{align}$$

Differentiating $k-1$ times yields $$\dfrac{\mathrm d^{k-1}}{\mathrm dz^{k-1}}\left(z\mapsto (z-z_0)^k\varphi(z)\right)(z)=(k-1)!c_{-1}+k!c_0(z-z_0)+\ldots.$$

Thus, evaluating the above at $z_0$ gives you the residue since everything on the RHS except for $(k-1)!c_{-1}$ goes astray. However $\varphi$ is not defined at $z_0$, so you can't really evaluate at $z_0$, that's OK, take the limit at $z_0$ which exists by definition of pole of order $k$.

This justifies the formula $$\text{Res}(\varphi, z_0)=\dfrac 1{(k-1)!}\lim \limits_{z\to z_0}\left[\dfrac{\mathrm d^{k-1}}{\mathrm dz^{k-1}}\left(z\mapsto (z-z_0)^k\varphi(z)\right)(z)\right].$$

For this particular example, I think it's better to compute $c_{-1}$ by using the series expansion.

Set $g\colon \mathbb C\to \mathbb C, z\mapsto \cos(z)-1, S=\{2k\pi i\colon k\in \mathbb Z\}, h\colon \mathbb C\setminus S\to \mathbb C, z\mapsto \dfrac1{\left(e^z-1\right)^2}$, and $f\colon \mathbb C\setminus S\to \mathbb C, z\mapsto g(z)h(z)$.

Let $k\in \mathbb Z$ be such that $k\neq0$ and write $s=2k\pi i$.

'Around' $s$ each of the functions above is (holomorphic, hence) analytic.

Therefore, for some sequences $(a_n)_{n\in\mathbb Z}, (b_n)_{n\in \mathbb Z}, (c_n)_{n\in \mathbb Z}$ and for all $z$ in a sufficiently small punctured disc centered at $s$ one has

$$g(z)=\sum \limits_{n\in \mathbb Z}\left(a_n(z-s)^n\right),$$ $$h(z)=\sum \limits_{n\in \mathbb Z}\left(b_n(z-s)^n\right),$$ $$f(z)=\sum \limits_{n\in \mathbb Z}\left(c_n(z-s)^n\right).$$

You've determined that $s$ is a pole of order $2$ of $f$. The same happens to $h$ since $\lim \limits_{z\to s}\left(\left|(z-s)h(z)\right|\right)=\infty$ and $$\lim \limits_{z\to s}\left((z-s)^2h(z)\right)=\lim \limits_{z\to s}\left(\dfrac{2(z-s)}{2(e^z-1)e^z}\right)=\lim \limits_{z\to s}\left(\dfrac{1}{2e^{2z}-e^z}\right)=1\neq 0 \tag{$\huge ☺$}$$

Thus $$\forall n\in \mathbb Z(n<0\implies a_n=0),$$ $$\forall n\in \mathbb Z(n<2\implies b_n=0),$$ $$\forall n\in \mathbb Z(n<2\implies c_n=0).$$

Therefore $$\sum \limits_{n=-2}^\infty\left(c_n(z-s)^n\right)=f(z)=g(z)h(z)=\sum \limits_{n=0}^\infty\left(a_n(z-s)^n\right)\sum \limits_{n=-2}^\infty\left(b_n(z-s)^n\right).$$

Recall that the goal is to find $c_{-1}$.

In particular and more pictorially one has $$\begin{align} g(z)h(z)&=\left(a_0+a_1(z-s)+\ldots\right)\left(b_{-2}(z-s)^{-2}+b_{-1}(z-s)^{-1}+\ldots\right)\\ &=\left(a_0b_{-2}(z-s)^{-2}+(a_0b_{-1}+a_1b_{-2})(z-s)^{-1}+\ldots\right). \end{align}$$

Consequently $c_{-1}=a_0b_{-1}+a_1b_{-2}$.

Because $g$ is analytic, $a_0$ and $a_1$ are just the coefficients of the taylor series of $g$ around $s$, that is, $$a_0=f(s)=\dfrac{e^{i\cdot 2k\pi i}+e^{-2k\pi i\cdot i}}{2}-1=\cosh(2k\pi)-1$$ and $$a_1=f'(s)=-\sin(s)=-\dfrac{e^{2k\pi i\cdot i}-e^{-2k\pi i\cdot i}}{2i}=\dfrac{e^{2k\pi}-e^{-2k\pi}}{2i}=-i\sinh(2k\pi).$$

Where as $\displaystyle b_{-2}=\dfrac{1}{2\pi i}\int _\gamma \dfrac{h(z)}{(z-s)^{-1}}\mathrm dz$ and $\displaystyle b_{-1}=\dfrac{1}{2\pi i}\int _\gamma h(z)\,\mathrm dz$, where $\gamma$ is an appropriate positively parametrized circle, centered at $s$, going around exactly once.

The integrals can be computed using the residue theorem.

By choosing $\gamma$ with a small enough radius, the only singularity inside the $\text{im}(\gamma)$ is $s$.

One gets $$\int _\gamma \dfrac{z-s}{(e^z-1)^2}\mathrm dz=2\pi i\text{Res}\left(z\mapsto \dfrac{z-s}{(e^z-1)^2}, s\right).$$

The formula given in the first part of this answer, together with $\huge ☺$, yields $\text{Res}\left(z\mapsto \dfrac{z-s}{(e^z-1)^2}, s\right)=1$, thus $b_{-2}=1$.

For the other, since $s$ is a pole of order $2$, one gets $$\begin{align} \int _\gamma h(z)\,\mathrm dz=\int _\gamma \dfrac{1}{(e^z-1)^2}\mathrm dz&=2\pi i\text{Res}(h,s)\\ &=2\pi i\lim \limits_{z\to s}\left[\dfrac{\mathrm d}{\mathrm dz}\left(z\mapsto (z-s)^2h(z)\right)(z)\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{\mathrm d}{\mathrm dz}\left(w\mapsto \dfrac{w^2}{(e^{w+2k\pi i}-1)^2}\right)(z)\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{\mathrm d}{\mathrm dz}\left(w\mapsto \dfrac{w^2}{(e^{w}-1)^2}\right)(w)\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{2w(e^w-1)^2-2w^2(e^w-1)e^w}{(e^w-1)^4}\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{2w(e^{w}-1)-2w^2e^w}{(e^{w}-1)^3}\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{2w\left(w+\frac{w^2}2+\ldots\right)-2w^2(1+w+\ldots)}{(w+\ldots)^3}\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{2w^2+w^3-2w^2-2w^3+\ldots}{w^3+\ldots}\right]\\ &=2\pi i\lim \limits_{w\to 0}\left[\dfrac{-w^3+\ldots}{w^3+\ldots}\right]\\ &=-2\pi i, \end{align}$$

therefore $b_{-1}=-1$.

Thus $\text{Res}(f,s)=c_{-1}=1-\cosh(2k\pi)-i\sinh(2k\pi)$. This agrees with wolfram alpha.

  • 2
    $\begingroup$ Strange. In my experience, for small orders, the Laurent expansion is usually the easiest way. $\endgroup$ Oct 22 '14 at 13:21

Evaluate $\frac{d}{dz}\left[(z-2\pi k i)^2 f(z)\right]$ at $2\pi k i$. (There is a general rule for finding the residue of a function at a pole by computing derivatives. Could get tedious for high order poles but for order 2, it is not too bad.)


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