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Understanding Cauchy's Integral Theorem

  1. Oct 9, 2007 #1
    The Cauchy Integral Theorem says that given f(z), an analytic function in a simply connected domain D and C, a Jordan curve contained in D:


    The words "simply connected" are important because if the closed contour encloses points where the function is not analytic it may have a value other than zero.

    There are a few things that confuse me here:

    1. Is a branch point always a point where the function is not analytic?
    2. Is a point where the function is undefined, that is not a branch point always a point where the function is not analytic?
    3. Intuitively, why should the integral be zero? Is it because the line integral of a complex function is summing up direction vectors at every point in the curve, and when you return to where you started on a closed curve you have essentially "not moved" so the sum of the direction vectors is zero?
    4. When a closed contour's integral has a value other than zero, is this ever related to the fact that you went over a branch cut? If not why will this happen?
  2. jcsd
  3. Oct 9, 2007 #2
    1) Yes, at a branch point the function is not analytic.

    2) Yes, if the function is not defined at a point then you cannot say that it is analytic at that point. It could be that you could define the function at that point in such a way that it would become analytical, but as long as you don't, it is not analytical.

    3) What is the integral of a function along the real axis? If F is the antiderivative it is the difference of F at the end and starting point. You can think of the contour integral being zero as saying that an integral from z1 to z2 will yield the same value irrespective of which path is used (because the difference of the integral along two paths is a contour integral along a closed path which is zero). Now, fix a point z0 and define a function F(z) as some path integral from z0 to z. Now, what do you think is the derivative of F? :smile:

    4) It will be more closely related to the fact that you went over a brach cut of F, not f. Example: evaluate the contour integral over the unit circle of the function [tex]\frac{1}{z}[/tex]. If we naively try to evaluate the integral using a antiderivative as I explained above you would get the difference of Log(z) at end and starting point. Now log has a brach cut singularity so the value jumps by [tex]2\pi i[/tex].
  4. Oct 9, 2007 #3
    f(x) right?

    Are you saying that the entire complex plane is like a big real number line in 2D? That alomst makes sense, though I'm still having a hard time seeing why the integrals over paths that "look like differnt lengths" give the same result.

    I mean, on the real line there is only one way to get fron 2.5 to -3.

    So is it that there are many ways to get from 2 + 3i to 3 - i? ... I'm still not getting this...
  5. Oct 9, 2007 #4
    Yes, the derivative is f, you could try to prove it...

    You can take any arbitrary path from 2 + 3i to 3 - i (not just a straight line) and the integral will be the same (if the function is analytic). It's just like the case of a conservative force field in physics, where the integral of the inner product of the force with the path length element along a path equals the potential energy difference at the end and starting points and is thus path independent.
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