- #1

Old Person

- 31

- 10

- 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.

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?

Thank you for your time and any advice.

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?

Thank you for your time and any advice.