# Cauchy,s theorem in Real plane?

1. Aug 11, 2004

### eljose79

i suppose all of us know the famous Cauchy,s formula

Int(c)f(z)dz/(z-a)=(2pi)if(a)

but could be applied the same to real plane,in fact let be the Integral over the closed path f(x,y)dxdy where f(x,y)=Gradient(g) then the integral in R^2

Int(C)f(x,y)dxdy/[x-a]=?

where [x-a] means sqrt[(x-a)^2+(y-b)^2] could be Cauchy,s theorem be applied in the real plane?..thanks.

2. Aug 11, 2004

### mathwonk

The reason Cauchy's theorem works, stated in real terms is that the complex form dz/(z-a), written out as a real form, has an exact part which gives zero integral around a closed path, plus if(a) times the real form "dtheta" =

-(y-d)/((x-c)^2+(y-d)^2)dx + (x-c)/((x-c)^2 + (y-d)^2) dy,

where a = c + di are the real and imaginary parts of the complex number a.

So I think you do not want to divide by sqrt((x-c)^2+(y-d)^2), rather you may need (x-c)^2+(y-d)^2, as in the denominator of dtheta, to get something interesting.

In fact dtheta, i.e. imaginary part of dz/z, is pretty much the only interesting such integral.

try reading something about deRham cohomology, if you can find anything understandable, maybe Differential topology by Guillemin and Pollack.

here is a theorem: among all possible 1 forms, i.e. expressions Pdx +Qdy, defined in the complement of the origin, among those that have curl equal to zero, i.e. such that dQ/dx = dP/dy, (partial derivatives), essentially the only one that is not the gradient of a function, is dtheta.

I.e. given any such 1 form, there is a function f and a number c such that

Pdx + Qdy = gradf + c(dtheta). Then the integral of both sides over any path going simply once around the origin, equals 2pi c.

I.e. as far as integrating over a circle around the origin, the gradf part gives zero, and hence Pdx +Qdy is equivalent to c(dtheta).

Last edited: Aug 11, 2004