Do you know how to do this integral please?

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SUMMARY

The integral of cosh(z)/(z^2 - z) over various closed contours can be evaluated using Cauchy's integral formula. Specifically, for the contour |z|=2, the integral evaluates to 0 due to the absence of singularities within the contour. For |z-i|=1/2 and |z-1|=1/2, the integrals can be computed by identifying the poles at z=0 and z=1, respectively, and applying the residue theorem. The transformation of cosh(z) into its exponential form is essential for simplifying the integrand.

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blueyellow
What is the value of

closed integral (C) of [cosh z/(z^2 -z)] dz

if C is
a) the circle |z|=2
b)the circle |z-i|=1/2
c)the circle |z-1|=1/2

i've tried reading up on cauchy's integral formulae in textbooks and my notes but i couldn't find an example which was similar. what do i do? do i hav to first turn the cosh z into a formula involving e's?
 
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is the whole z thing in cos ?

if that is so then take z common cancel it
you'll be left with 1/(Z-1)
 

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