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Cauchy,s theorem in Real plane?

  1. Aug 11, 2004 #1
    i suppose all of us know the famous Cauchy,s formula


    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


    where [x-a] means sqrt[(x-a)^2+(y-b)^2] could be Cauchy,s theorem be applied in the real plane?..thanks.
  2. jcsd
  3. Aug 11, 2004 #2


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    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
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