# Calculate the residue of 1/Cosh(pi.z) at z = i/2 (complex analysis)

• I
• Old Person
In summary, the question asks for a method to determine the residue of a function with a singular point at z=i/2. One possible approach is to use the formula Res (f(z); z_0) = lim(z->z_0) [(z-z_0) * f(z)] and the fact that the point of interest is a pole of order 1. Another method is to obtain the Laurent series for the function and use the coefficient of the term with the highest negative power of z as the residue. However, this method may not work for functions with isolated singular points that cannot be described as a pole.
Old Person
TL;DR Summary
Calculate the residue of 1/Cosh(pi.z) at z = i/2.
I'm not sure if this should be in the calculus section or the anlaysis section. It's complex analysis related to integration around a contour.

Can someone suggest a method to determine the residue of f(z) = ## \frac{1}{Cosh ( \pi z) } ## at the singular point z = i/2.

Background:
This was part of an exam question and the model answers are already available.
In the model answers it is just assumed we have a simple pole (pole of order 1) at z=i/2 and the usual formula then jumps in:
Res (f(z) ; i/2) = ## \lim\limits_{z \rightarrow i/2} \, [ (z-i/2) . f(z) ]##
BUT no proof is given that we did only have a simple pole there. If that's easily shown, great, otherwise would it be easier to just directly obtain the Laurent series or obtain the Residue by some other method?

FactChecker
I used ##Res_z\left(\dfrac{1}{f}\right)=\dfrac{1}{f'(z)}## from
https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/
and got
\begin{align*}
Res_z\left(\dfrac{1}{f}\right)&=\dfrac{1}{f'(z)}=\dfrac{1}{(\cosh \pi z)'}\\&=\dfrac{1}{\pi \sinh \pi z}=\dfrac{1}{\pi \sinh(i \pi/2)}=\dfrac{1}{\pi i}=-\dfrac{i}{\pi}
\end{align*}
if I made no mistakes by looking up the complex arguments of ##\sinh## on Wikipedia.

https://www.wolframalpha.com/input?i=residue+(1/cosh(pi+z);+i+/2)=

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Old Person
Hi and thank you ever so much for your time and speed of reply @fresh_42

The answer is OK in that it does agree with the model answer. However, I'm not sure it was a valid method.

I've just been looking over the insight article you suggested. It states, just above the result : Say Z_m is a zero of order m and.... .... then gives the formula you've used for z_1 only:

## Res_{z_1}\left(\dfrac{1}{f}\right) =\dfrac{1}{f'(z_1)} ##

So it looks like you must already know that the point of interest, z, was a pole of order 1 to use that result. I'm not sure, I didn't write that article - but that's what it looks like. Just to be clear then, it may be right, it's just that we still seem to need to establish that there was a simple pole (of order 1) there.

As for using Wolfram Aplha approach - as it happens I already tried it. Sadly it spits out answers but no explanation or method. Did it also just assume a pole of order 1 was there?

Yes, the order of the pole is needed beforehand, and I guess WA determined it, too, as part of its calculations.

Old Person
##\cosh (i \pi/2 )=\dfrac{1}{2}\left(e^{i \pi/2}+e^{-i \pi/2}\right)=\dfrac{1}{2}\left(i - i\right)=0.## Therefore ##\cosh(z)=(z-i \pi/2)\cdot g(z).##

Assume that ##z=i\pi/2## is of higher order than ##1,## i.e. ##\cosh(z)=(z-i \pi/2)^2\cdot h(z).## Then
\begin{align*}
\dfrac{d}{dz}\cosh(z)&=\sinh(z)=2(z-i \pi/2)\cdot h(z)+ (z-i \pi/2)^2\cdot h'(z)
\end{align*}
The RHS is zero at ##z=i \pi/2## but ##\sinh(i \pi/2)=\dfrac{1}{2}\left(e^{i \pi/2}-e^{-i \pi/2}\right)=\dfrac{i-(-i)}{2}=i\neq 0.##

This proves that the zero ## i \pi/2## of ##\cosh(z)## is of order one.

weirdoguy and hutchphd
You cna get the Laurent series for $\operatorname{sech}(\pi z)$ about $i/2$ by setting $t = z - \frac i2$ and then $$\begin{split} \frac{1}{\cosh(i\pi/2 + \pi t)} &= \frac{1}{\cosh(i\pi/2)\cosh \pi t + \sinh(i\pi/2) \sinh \pi t} \\ &= \frac{1}{i\sinh \pi t} \\ &= \frac{1}{i\pi t} \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} \\ &= \frac{1}{i\pi t}\left( 1 - \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right) + \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right)^2 + \dots \right) \\ &= \frac{-i}{\pi(z - \frac i2)} + \sum_{n=0}^\infty a_n(z - \tfrac i2)^n. \end{split}$$ But it is enough to note that $$\frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right)$$ and the second factor is analytic at $t = 0$.

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Old Person and SammyS
Hi and thank you for even more replies.

@fresh_42 ---> That's a good answer. I think there is one minor issue but it is minor. It won't be a problem unless 1/Cosh z is extremely unusual.

fresh_42 said:
Assume that z=iπ/2 is of higher order than 1, i.e. cosh⁡(z)=(z−iπ/2)2⋅h(z). Then....

You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).

Example ##e^{1/z} ## = Exp (1/z) = ## \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{z^n} ## = a Laurent series about z=0. There is no finite n where the thing can be written as F(z) / zn (with F(z) analytic in a small neighbourhood of z=0).
If the singular point cannot be described as a pole then you can't apply the differentiation trick because you won't have the ability to write ## \cosh(z)=(z-i \pi/2)^N\cdot g(z) ## (with g(z) analytic), for any finite N.
- - - - - - - - - -

@pasmith , that may actually work. In some lines of the work on those series, you needed the series to be absolutely convergent to be sure you could re-arrange and re-shuffle the terms. However, I think you've got that. I wouldn't have had time to check all these details in the exam - but I do now and they they do seem to be ok.

Where you write: ..it is enough to note.... ## \frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right) ##

I think I can see what you are saying. This bit ## \left( \frac{\pi t}{\sinh \pi t} \right) ## = ## \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} ## from earlier in your post and the RHS is seen to be analytic at t=0. So that is enough to establish that you have a simple pole.

- - - - - - - - - - -
Overall, thank you to everyone. I'm happy with the answers. It's been great to see that it could be done (given time) but also just to have a couple of people acknowledge that there was a need to establish the nature of the singular point (or directly look at the Laurent series). As I mentioned before - in the model answers, there are no marks for proving that or even indicating that you should care. It's clear the person who set the exam just expected you to assume you had a simple pole there.

Old Person said:
You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).
I actually haven't said anything about poles. I only have analyzed the point ##z=\mathrm{i} \pi /2## of the analytical function ##z\longmapsto \cosh(z).##

I showed that it is a zero of order one. That makes its inverse ##z \longmapsto \dfrac{1}{\cosh(z)}## having a pole at ##z=\mathrm{i} \pi/2## of order one. It's not isolated since the zero wasn't isolated.

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