Compute the complex integral 1/(2pi*i) oint_C ((ze^z)/(z-i)) dz . z is a complex number, C is a circle or radius 2 centered at z=0 in the complex plane oriented counterclockwise.

This is an integral of the form oint_C f(z)/(z-a) dz , where f(z)=ze^z and a=i in this case. If this meets the conditions of the Cauchy Integral Theorem we can use the Cauchy Integral Formula.
oint_c f(z)/(z-a) dz =2pi i*f(a)
f(z) must be a holomorphic function. ze^z is indeed an holomorphic as its infinity differentiable everywhere inside the boundary C .
a must be inside the bound curve C . i is inside a circle of radius 2 centered at z=0 .
Therefore by the Cauchy Integral Formula:
1/(2pi*i) [oint_C (f(z))/(z-a) dz]=1/(2pi*i) [oint_(|z|=2) (ze^z)/(z-i) dz]=1/(2pi i)*[2pi i*f(i)]=f(i)=ie^i
1/(2pi*i) oint_(|z|=2) (ze^z)/(z-i) dz=ie^i
http://stat.math.uregina.ca/~kozdron/Teaching/Regina/312Fall12/Handouts/312_lecture_notes_F12_Part2.pdf