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Copmlex Integration

  1. Feb 14, 2009 #1
    If [itex]\Gamma[/itex] is the closed path as follows:
    from [itex]\delta[/itex] to R along the positive real axis then around the semi circle of radius -R on the upper half plane to -R on the negative real axis then along the negative real axis to [itex]-\delta[/itex] then around the semi circle of radius [itex]\delta[/itex] in the upper half plane and back to [itex]\delta[/itex] on the positive real axis.

    If [itex]f(z)=\frac{1-e^{iz}}{z^2}[/itex], explain why [itex]\int_{\Gamma} f(z) dz=0[/itex]?

    I was thinking of finding a domain,U, for f that was also a semi-annulus in the upper half plane but with outer radius much greater than R (say 1000R) and inner radius infinitesimaly small. Then U is star shaped if we pick the star centre at (0,1000R). [itex]f:U \rightarrow \mathbb{C}[/itex] will be holomorphic and so we can use Cauchy's Theorem to give the answer - but it doesn't seem like a very rigorous definition of U.

    can anyone advise me?
     
  2. jcsd
  3. Feb 14, 2009 #2

    Dick

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    Isn't f(z) already holomorphic on the domain in the contour? Why do you want to define another contour?
     
  4. Feb 14, 2009 #3

    HallsofIvy

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    [tex]f(z)= \frac{1- e^{iz}}{z^2}[/tex] is analytic everywhere except z= 0 and that is outside the original path. Changing the diameter of the the two circles wont' change that.

    Is the problem that the interior is not "star shaped" (there exist at least one point such that the straight line from that point to every other point is in the region)?

    First, no matter how large you make one circle or how small you make the other, it will never be "star shaped". Second, Cauchy's Theorem does not apply only to star shaped regions.
     
  5. Feb 14, 2009 #4
    the only part of my notes i can find that says cauchy's theorem doesn't need a star shaped domain is called the "contour moving technique"? but that doesn't seem much use here...
     
  6. Feb 15, 2009 #5
    anybody?
     
  7. Feb 15, 2009 #6

    Dick

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    What's the 'contour moving technique'?
     
  8. Feb 15, 2009 #7

    HallsofIvy

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    It's impossible to know "where you are" in your course and so we don't know what you have available to use. But it is true that as long as you move a contour "continuously" without crossing any discontinuities of the function, the integral around that contour remains the same. The integral around any "Jordan curve" (basically no self-intersections) is 0 as long as the function is analytic inside and on the curve.
     
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