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What methods can I use to determine the residues of $\frac{z}{\sinh(kz)}$? Singularities occur at $z=\frac{i n \pi}{k}$ for $k \neq 0$ and $n$ an element of the integers.

I've attempted a series derivation of the reside:

$$\frac{z}{\sinh(kz)}= \frac{1}{1+\frac{1}{6}z^2+\frac{1}{120}z^4}$$

However, I don't understand the arithmetic required to 'get the series to the numerator', which will allow me to determine the coefficient of the $\frac{1}{z}$ term. The problem I am having is discussed here, yet I still don't understand it... How is this series in denominator converted to a series in numerator?

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Tunk
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  • I would’ve started with the definition of $\sinh$ as $(e^x-e^{-x})/2$ and noticed that its zeros are all on the imaginary axis. – Lubin Apr 16 '16 at 02:05
  • @Lubin Yes I've noticed that, however, how does that help me in calculating the residues at these particular points? – Tunk Apr 16 '16 at 02:06
  • To tell the truth, I stopped there because I didn’t know how to go beyond it. You’ll have to wait for somebody who knows what (s)he’s doing. – Lubin Apr 16 '16 at 02:09
  • @Lubin Got it - thanks anyways. I've edited the problem to include additional information regarding the singularities. – Tunk Apr 16 '16 at 02:11
  • Your series, besides omitting $k$, will only tell you what’s going on at $0$. Supposeing $k=1$, though, why don’t you make a substitution $z\mapsto z - n\pi i$, when you can now use the series expansion at $0$. – Lubin Apr 16 '16 at 02:13
  • @Lubin Like so: $\sinh(z)=\sinh(z'-n \pi i)=(-1)^n\sinh(z')$? – Tunk Apr 16 '16 at 02:22
  • I think I can write it up. – Lubin Apr 16 '16 at 02:27

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I’m going to make things easy for myself by taking the case $k=1$ only.
The zeros of the denominator are at $n\pi i$ for $n\ne0$, and instead of trying to expand the original function in the neighborhood of $n\pi i$, in powers of $z-n\pi i$, I’ll get that expansion at zero by looking at the expansion around zero of $f(z-n\pi i)$.

We have \begin{align} f(z-n\pi i)&=\frac{z-n\pi i}{\sinh(z-n\pi i)}\\ &=\frac{z-n\pi i}{(-1)^n\sinh z} =\frac{(-1)^nz}{\sinh z}+\frac{(-1)^{n+1}n\pi i}{\sinh z}\,. \end{align} Now, the $z/\sinh z$-part has no contribution to the residue, so we can forget that. However, the residue of $1/\sinh z$ is easily determined, since $(1+x^2/6+x^4/120+\cdots)^{-1}$ is a series that starts out in the shape $1 - x^2/6 + 7x^4/360-\cdots$, as you can easily check by long division of power series. (It’s just like long division of polynomials, but you write the terms in ascending order of degree instead of descending.) Actually, of course you only need to know that the series starts out with constant term $1$, because then you see that $1/\sinh z$ has Laurent series $z^{-1}-z/6+\cdots$, and it’s that $z^{-1}$ term that makes the only contribution to the residue.

I think you can do the rest.

Lubin
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