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Branch cut contradiction

  1. Apr 3, 2006 #1
    In complex analysis we say that for fn's like lnz we apply a branch cut along positive x-axis to make sure it's single valued. i.e restrict theta s.t 0<=theta<2Pi but we never allow theta to equal 2Pi as this would make lnz take on 2nd value.
    Let us integrate around a contour which goes from origin to x= infinity, then goes anticlockwise around a circle of infinite radius back to the positive x-axis at plus infinity, then we go back to the origin along the x-axis, and then go clockwise around a circle of zero radius, hence avoiding crossing the branch cut. we say that theta = 0 as we go from the origin out to x=infinity, and then we have to say (in order to get the answer right) that theta = 2pi when we go from x-infinity back to the origin. But surely this is a contradiction, since we assumed that theta could not equal 2 Pi?
     
  2. jcsd
  3. Apr 3, 2006 #2

    JasonRox

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    Can you better describe this contour?

    Try not to lump everything you say in one sentence or even paragraph.
     
  4. Apr 3, 2006 #3

    shmoe

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    You are never integrating over circles with "infinite" or "zero" radius. Your contours are compact and avoid the origin.

    You can use a "keyhole" contour to avoid the real axis and clear up your concerns (see diagram near the bottom of http://en.wikipedia.org/wiki/Residue_calculus ). The horizontal segments lie strictly above and below the positive real axis, and get closer and closer to it as your contour grows. As this segment below the real axis is approaching it, the argument of log is going to 2*Pi and the integral over this segment will approach the integral over the real axis of this different choice of branch cut.
     
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