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Deforming Contour of complex function

  1. Feb 4, 2009 #1
    First, thank you for reading this.

    I've got a complex function, F(z), which is assumed to be analytic, and I know it's values along a contour in the complex plane. Say, for simplicity, that contour is known parametrically as x(t) = cos(t), y(t) = sin(t), 0 < t< pi, thus I know F(x(t) + i*y(t)), and possibly dF/dt.

    What I want to do is obtain F(x), that is what is F along the real axis?

    What I have done so far is to assume that F obeys the Cauchy-Riemann equations, then if
    F(x,y) = u(x,y) + i*v(x,y), I can get du/dy and dv/dy from the information in dF/dt (sets up a nice little matrix problem). Then I can take an integration step towards the real axis to get F at a new contour (e.g. y(t) = 0.99sin(t), x(t) = cos(t)), I can then find dF/dt along this contour (sadly, only through a finite difference formula, or interpolation) and then find du/dy and dv/dy again, take a new step, etc.. until y(t) = 0. It works pretty well for simple test cases, such as, F(z) = z^n, etc. However, it seems to be a little unstable as well when F gets a little more complicated, that is errors tend to build and sometimes the function "explodes" before I get to the real axis.

    What I would like to know is if this has been done (in a better way) by someone else, and where I could go about looking. My google searches haven't resulted in much and I don't know my way around the math journals. Any suggestions?

  2. jcsd
  3. Feb 4, 2009 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    This can be done. If F(z) is analytic and defined over some contour of nonzero length, it should be possible to analytically continue F to the whole complex plane (in a unique way).

    The Kramers-Kronig relations demonstrate one example. If F is defined on either the real axis or the imaginary axis, the entire function can be found by doing some integrals.
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