If there is a singularity in your region, then no, it is not a simple closed contour. You will need to evaluate residues.
Hi.
Evaluate the integral with respect to C, C={ z: lz+2l = 3}, Integrate cos(pi*z)/( ((z+4)^2)*(z-2).dz
I understand that from the denominator the value -4 lies inside C but the point 2 does not.
Am i right in thinking then f(z)=(cos(pi*z)) / ((z-2)^2) is analytic on R and C is a simple closed contour in R. Hence by Cauchy's First Derivative Formula......
....and this is where I'm unsure.
Any suggestions?
Shoni