Is Cauchy-Riemann a necessary or sufficient condition?
However, it is possible to have an integral that doesn't satisfy Cauchy-Riemann and yet the method of residues still works! To see an example of such an integral, see the Appendix of this document:
The questions are:
a) Is this type of exception generally known?
b) Can you explain what is going on and why contour integration by residues still works?
c) Is it possible to define a class of integrals that are solvable by residues, and yet do not satisfy Cauchy-Riemann?