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Independence of path

  1. Mar 29, 2006 #1
    I am trying to understand how the Cauchy-Goursat theorem of complex analysis differs from the usual conditions for independence of path in real vector calculus.

    My complex analysis textbook emphasizes that the Cauchy-Goursat theorem is true even if the function we are integrating does not have continuous first partials and that this differs from the real case.

    Looking over the proof in my textbook, I don't really see anything being done that couldn't be done in R^2. So why do we need the first partials to be continuous when we integrate over a closed curve in R^2 but not when we integrate over a closed contour in C?
  2. jcsd
  3. Mar 29, 2006 #2


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    What do you call the partial derivative of a complex function of complex variables? I'm assuming you're refering to the partial derivatives of the real and imaginary parts.

    What version of the Cauchy-Goursat theorem are you refering to?

    Because I know it to be

    'f holomorphic on an open U and C a contractible Jordan curve in U. Then the integral of f over C is 0'.

    Here, the fact that f is holomorphic implies by the Cauchy-Riemann thm* that the partial derivatives are in fact continuous.

    * see the 'Formulation' paragraph here: http://en.wikipedia.org/wiki/Cauchy-Riemann
  4. Mar 29, 2006 #3
    Yes, I am referring to the partial derivitives of the real and imaginary parts.
    Also, holomorphic and analytic are the same thing, right quasar?

    The version of the Cauchy-Goursat theorem I am referring to is:

    If a function f is analytic at all points interior to and on a simple closed contour C, then the integral over C of f is 0.

    The condition of analyticity does seem to imply that the partials are continuous...

    I think perhaps what my textbook means by saying that the first partials need not be continuous is that the theorem can be proven without assuming that fact. This then allows us to prove using Cauchy-Goursat the fact that analyticity implies that the first partials are continuous...
  5. Mar 29, 2006 #4


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    the two theorems have the same conclusion but different hypotheses.

    one (over the reals) assumes continuous partials, and the other (over the complexes) only assumes the partials exist and satisfy the cauchy riemann equations. but in fact it is a complex theorem that any solution of the cauchy riemann equations is actually analytic, hence has not only continuous partials, but infinitely many continuous partials.
  6. Mar 29, 2006 #5
    And to prove that theorem you would need to use Cauchy-Goursat right?

    So the real important difference between the real and complex cases must be in the fact that the Cauchy-Goursat theorem doesn't need to assume the first partials are continuous. So my question is, what property of complex numbers is invoked in the proof of Cauchy-Goursat that allows us to drop the extra hypothesis?

    Looking at the proof I don't really see anything that doesn't look like it could be done in the real case just as well... at least nothing jumps out at me like "oh, thats why we can't do this in R^2."
  7. Mar 29, 2006 #6
    Affine Plane vs. Complex Plane

    I will just throw out some ideas here:

    For starters, vectors in [itex]R^2[/itex] inhabit the affine plane whereas the complex numbers [itex]C[/itex] inhabit the complex plane and these two spaces are a little conceptually different even though mappings from one to the other are quite simple to do.

    Vectors in [itex]R^2[/itex] express directed magnitudes with absolute direction whereas [itex]C[/itex] express directed magnitudes with relative direction. Multiplication by a complex number results in a dilation and rotation of a vector. The magnitude and direction of the new vector is relative to the magnitude and direction of the old one. To see this clearly, we can use the following isomorphism between [itex]C[/itex] and [itex]2 \times 2[/itex] matrices:

    [tex]\rho e^{i \theta} \longleftrightarrow \left(\begin{array}{cc} \rho\cos\theta & -\rho\sin\theta \\ \rho\sin\theta & \rho\cos\theta \end{array}\right)[/tex]


    [tex]u + i v \longleftrightarrow \left(\begin{array}{cc} u & -v \\ v & u \end{array}\right)[/tex]

    Conformal maps have the property that they conserve angles. Holomorphic maps [itex]C \rightarrow C[/itex] are conformal by virtue of the fact that a given complex number rotates any vector by the same amount as any other vector. At any internal point where it is holomorphic we can find an associated map [itex]R^2 \rightarrow R^2[/itex] and see how our map transforms vectors in some small neighborhood around the point. For very small neighborhoods, all vectors should be transformed the same way with only a very small difference as we go along a curve through this point - and this should be independent of the particular path we take.

    In general, integrals of maps [itex]T: R^2 \rightarrow R^2[/itex] will not be path-independent. It seems the requirement that its partials be continuous is necessary to establish that the integrals are path-independent.
    Last edited: Mar 29, 2006
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