I would like to know if there is a quick way of computing the residues of $$f(z) = \frac{\cot \pi z}{z(z+1)}$$at the points $z = 0$ and $z = -1$. They are double poles.

Expanding this in Laurent series I have: \begin{align} \frac{\cot \pi z}{z(z+1)} &= \frac{1}{z}\left(\frac{1}{\pi z}-\frac{\pi z}{3}-\frac{\pi^3z^3}{45}+\cdots\right)(1-z+z^2-z^3+z^4+\cdots) \\ &= \frac{1}{\pi z^2}-\frac{1}{\pi z}+\frac{1}{\pi}-\frac{\pi}{3}+\cdots\end{align} which gives: $${\rm Res}\left(\frac{\cot \pi z}{z(z+1)}, 0\right) = -\frac 1\pi.$$ My calculation above seemed quick, but computing Laurent series for $\cot$ is a pain. Also, Wolfram Alpha gives: $${\rm Res}\left(\frac{\cot \pi z}{z(z+1)},-1\right) = -\frac 1\pi,$$too, which makes me think that there is some symmetry to be explored here. At first I thought about some symmetry around the $x = -1/2$ axis, and the following plot suggests that it might be the case. However, we don't have quite a straight line, so I don't know how to apply this.

enter image description here

up vote 3 down vote accepted

You can do this without doing much work as follows. You can write:

$$\frac{1}{z(z+1)} = \frac{1}{z} - \frac{1}{z+1}$$

Also, if we have some analytic function $f(z)$ with simple zeros at $z_k$, then its logarithmic derivative $g(z)$ will have poles of order 1 with residues equal to 1 at $z_k$. If we take $f(z) = \sin(\pi z)$ then the zeros are at integer $z$ and the logarithmic derivative is $g(z) = \pi\cot(\pi z)$ which up to the factor of $\pi$ is the cot function in your problem.

We also know that $g(z)$ is an odd function, the first term in the Laurent expansion is $\frac{1}{z}$, the next term is $\mathcal{O}(z)$. This implies that the residue at $z = 0$ won't get a contribution from the $\frac{1}{z}$ term, so only the $-\frac{1}{1+z}$ term can make a contribution. That contribution is equal to the value of this term at $z=0$ times the residue of the cot function at zero, which yields $-\frac{1}{\pi}$.

  • I think I understand. Mimicking your argument, I'd say that the residue at $-1$ would be the value of the term $\frac{1}{z}$ at $-1$ times the residue of the $\cot$ function at zero, which yields $-\frac{1}{\pi}$ too. Thanks, this works fine. Just to clarify, about stating that the final residue would be the value of something evaluated at the point times the residue of the initial function.. is there some result stated about this situation, or it was something along the lines of "just look at it"? I've never seen any result like this. +1 of course. – Ivo Terek Jul 19 '15 at 2:45
  • 1
    @IvoTerek If we write the function as f(z) g(z) and f(z) has a pole of order 1 at z = 0 while g(z) is analytic there then in the Laurent expansion of the product, to get the 1/z term, you must take the 1/z term from f(z) and the constant term from g(z), therefore the residue will be the residue of f(z) multiplied by g(0). – Count Iblis Jul 19 '15 at 2:51
  • @CountIblis If the Laurent expansion for $g$ has terms that are of "lower order" (e.g., $z^{-2}$, $z^{-3}$), then the higher order terms in $f$ will be implicated in the residues of $fg$. – Mark Viola Jul 19 '15 at 2:57
  • @Dr.MV Indeed, so we are using that cot only has simple poles. – Count Iblis Jul 19 '15 at 2:59
  • I carried out some computations. Check it out: $$\begin{align} {\rm Res}\left(\frac{\cot \pi z}{z(z+1)},-1\right) &= {\rm Res}\left(\frac{\cot \pi z}{z},-1\right) - {\rm Res}\left(\frac{\cot \pi z}{z+1},-1\right) \\ &= {\rm Res}(\cot \pi z, -1) \frac{1}{z}\bigg|_{z=-1} - {\rm Res}\left(\frac{\cot \pi z}{z+1},-1\right) \\ &= -\frac{1}{\pi} - {\rm Res}\left(\frac{\cot \pi z}{z+1},-1\right) . \end{align}$$Similar computation gave the result for the residue at $0$ because $\cot \pi z = \frac{1}{\pi z} + {\cal O}(z)$, but how can we get rid of that term above? I don't think a parity argument will – Ivo Terek Jul 19 '15 at 3:12

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