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`