# Is Cauchy-Riemann a necessary or sufficient condition?

When integrating using the method of residues, one needs to check if the integrand satisfies the Cauchy-Riemann equations first.

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?

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 :-).

## All Answers (9)

Sajad Jafari· Amirkabir University of TechnologySorry for the poor English of my answer. I may use some incorrect English words in this answer.

I can say my opinion with an example. Suppose you want to calculate integral of a complex variable function like “f(z) = f(x+iy) = U(x,y)+iV(x,y)” in the area D by residues method.

1. When you integrate using the method of residues, your function should be analytical expect for countable points in the area of your integral.

2. Beside some other conditions (Ux,Uy,Vx,Vy should be continues), if Cauchy-Riemann conditions are satisfied, f is analytical.

So Cauchy-Riemann is not a necessary condition, it is sufficient (if other conditions will be satisfied too)

Jan Sładkowski· University of Silesia in KatowiceIn the standard formulation of the of the theorem in question the function actually has to be holomorphic and Cauchy-Riemann is a weaker condition! (In, addition the de partial derrivatives have to be continuous.). But as I remember, in the proof one uses the Green theorem and Cauchy-Riemann eqs. I have to check if such a formula can be proven with weaker assumptions. There Looman-Menchoff theorems that give the weaker condition for a continuous function to be holomorphic- if I remember correctly the the CR condition has to hold only almost everywhere.

Luca Dimiccoli· Vrije Universiteit BrusselDemetris Christopoulos· National and Kapodistrian University of Athens(CR+Continuous Partial Derivatives) is sufficient condition.

Arkady Kholodenko· Clemson UniversityLuca Dimiccoli· Vrije Universiteit BrusselJan 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 :-).

Demetris Christopoulos· National and Kapodistrian University of AthensThe real functions u(x,y), v(x,y) satisfy the conditions C-R and they are harmonic functions.

The complex function f is not analytic at origin z=0.

The path integral of f over any curve that contains the origin (0,0) is -2*pi.

The Res(f,0)=i, thus Integral=2*pi*i*i=-2*pi.

That is because u, and v have not cintinuous partial derivatives at origin.

Rahul Dravid· Bhagwant UniversityCan you help by adding an answer?