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Analytic functions and integration

  1. Oct 18, 2011 #1
    I have an example I want to clearify,

    Let C be a semicircle from 2 to -2 which passes through 2i and let T be a semicircle from 2 to -2 which passes through -2i.

    If you take the integral of z^2 dz around both paths the is the same, as the function is analytic so the integral is independant of the path,

    However if you take the integral of the conjugate of z over these to paths you get

    [PLAIN]http://img6.imageshack.us/img6/7467/unledpcv.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 18, 2011 #2
    [tex] e^{-it} e^{it} = e^0 = 1 [/tex]
     
  4. Oct 18, 2011 #3
    Oh okay,

    So they're both i4pi, but like, what does it mean that the integrals are the same over different paths and the function is not analytic?
     
  5. Oct 18, 2011 #4
    You're integrating the same constant from 0..pi and 0..-pi, so they are negative to each other and will cancel each other out.
     
  6. Oct 18, 2011 #5
    "If f(z) is analytic in a simply connected region then the int from one point to another is independent of the path joining the two points in the domain."

    So here f(z) isn't analytic, is the only reason the integrals are the same the trivial one?

    Because zbar isnt analytic at all
     
  7. Oct 18, 2011 #6
    No dude, I'm trying to tell you that you made a mistake in your calculation, so the integrals are NOT the same. Check your calculation please! The first integral should give 4 pi i and the second one should give -4 pi i
     
  8. Oct 18, 2011 #7
    Oh right, phew,

    Thanks man

    hey could i ask a small question,

    For scalar functions,

    is the orientation of the normal vector relavent? the tu X tv,
    where h(u,v) is the parametrization, tu tv its partial derivatives
    because when you take the surface integral you take f(h(u,v))||tu X tv||dudv, so the orientation really doesnt matter right?

    like for vector fields its the dot product of the two so the orientation of the vector does matter
     
  9. Oct 18, 2011 #8
    Yes, that's correct, the orientation of tu x tv doesn't matter if the function is scalar. What matters, of course, is the relative orientation of tu and tv, but that's pretty obvious. You know that |tu x tv| du dv is the Jacobian determinant in case you have different coordinates (like cylindrical or spherical), right? In case you're dealing with a manifold, you have the Gramian determinant.
     
  10. Oct 18, 2011 #9
    Sweet thanks alot Leon
     
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