Finding the residues of $\frac{\cos z -1}{(e^z-1)^2}$.

Question

I've found the poles of $\frac{\cos z -1}{(e^z-1)^2}$ to be double poles at each $z_k = 2k\pi i$, where $k\in\mathbb{Z}$ and $k\neq 0$. (At $k=0$ this is a removable singularity instead.)

I have no idea how to find out the residues at each $z_k$ - I tried using the formula $\lim\limits_{z\rightarrow z_{0}}\dfrac{d}{dz}(z-z_{k})^{2}\dfrac{\cos z -1}{(e^{z}-1)^{2}}$, but I can't see how to rearrange that and find the limit.

Is there a better way to do this?

Answer 1

Note that\begin{align}\cos(z)-1&=\cos\bigl((z-2k\pi i)+2k\pi i)-1\\&=\cos(z-2k\pi i)\cosh(2k\pi)-1-\sin(z-2k\pi i)\sinh(2k\pi)i\\&=\cosh(2k\pi)-1-i\sinh(2k\pi)(z-2k\pi i)+\cdots\end{align}and that\begin{align}(e^z-1)^2&=(e^{z-2k\pi i}-1)^2\\&=(z-2\pi i)^2+(z-2\pi i)^3+\cdots\end{align}So, if$$\frac{\cos(z)-1}{(e^z-1)^2}=\frac{a_{-2}}{(z-2k\pi i)^2}+\frac{a_{-1}}{z-k\pi i}+a_0+\cdots,$$then you have$$\cos(z)-1=\cosh(2k\pi)-1-i\sinh(2k\pi)(z-2k\pi i)+\cdots$$and also\begin{align}\cos(z)-1&=\cosh(2k\pi)-1-i\sinh(2k\pi)(z-2k\pi i)+\cdots\\&=(e^z-1)^2\frac{\cos(z)-1}{(e^z-1)^2}\\&=\bigl((z-2\pi i)^2+(z-2\pi i)^3+\cdots\bigr)\left(\frac{a_{-2}}{(z-2k\pi i)^2}+\frac{a_{-1}}{z-k\pi i}+a_0+\cdots\right)\\&=\bigl(1+(z-2k\pi i)\bigr)\bigl(a_{-2}+a_{-1}(z-2k\pi i)+a_0(z-2k\pi i)^2+\cdots\bigr)\\&=a_{-2}+(a_{-2}+a_{-1})(z-2k\pi i)+\cdots\end{align}So, $a_{-2}=\cos(2k\pi)-1$, and the residue, which is $a_{-1}$ can now be obtained from the equality$$a_{-2}+a_{-1}=-i\sinh(2k\pi).$$In other words, the residue is equal to $1-\cosh(2k\pi)-i\sinh(2k\pi)$.

Answer 2

You have already found that $z=0$ is a removable singularity and the poles are $z_k=2kπi$ with $k\in\mathbb{Z}, k\neq 0$.

$(e^z-1)^2$ is analytic on $\mathbb{C}$, so it can be expanded in a Taylor series (converging to it at all $z\in\mathbb{C}$) about $z=z_k$:

\begin{align}(e^z-1)^2&=(z-z_k)^2+(z-z_k)^3+\cdots\\&=(z-z_k)^2\left[1+(z-z_k)+\cdots\right]\\&=(z-z_k)^2 g_k(z),\end{align}

with $g_k(z_k)=g'_k(z_k)=1.$

The residue of the function at $z=z_k$ is:

\begin{align}&\lim\limits_{z\rightarrow z_k}\frac{d}{dz}\frac{\cos z -1}{g_k(z)}\\&=\lim\limits_{z\rightarrow z_k}\frac{(-\sin z)g_k(z)+(1-\cos z)g'_k(z)}{[g_k(z)]^2} \\&=1-\cos (2k\pi i)-\sin (2k\pi i) \\&=1-\cosh (2k\pi)-i\sinh (2k\pi).\end{align}