What is the integral e^z/Z-1 dz where C is the circle|z| = 2?

Expert Solution

Given curve |Z|=2 represents a circle with radius of 2 units also our complex integral has a singularity at z=1 ,so it must not be analytic at this point also this point lies inside the contour which is nice otherwise it's integral would have been 0.

Let.h(z) be the integrand . Then h(z) = e^z /(z -1) . and z =1 liesinside it..More over e^z is analytic inside and on that circle. So by Cauchy’s integral formula

the integral over | z| = 2 is 2(pi) i x e^1 = 2(pi)e.

2iπe

Arjoma Karan answered 2 years ago

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