# 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:

http://www.eleceng.adelaide.edu.au/personal/dabbott/conference/UPN_abbott1996b.pdf

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?

## Popular Answers

Jan Sładkowski· University of Silesia in KatowiceYesterday, I promissed to check if the assumptions can be weaker. I don't think so because the crucial point is that the integral along the closed path must vanish what is guaranteed by holomorphity. I can imagine a non-holomorphic function with vanishing integral along such path that fulfils the the residue theorem formula but, I think, this would be only a numerical coincidence. But an example would be interesting :-).