Finding the poles and residues of a complex function $\frac{\cos(z)-1}{(e^z – 1)^2}$

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

Solutions Collecting From Web of "Finding the poles and residues of a complex function $\frac{\cos(z)-1}{(e^z – 1)^2}$"

"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.

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.)