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Dirac delta function on the complex plane?

  1. Apr 16, 2003 #1
    Supposedly,

    ∫ ez*(z - z0)f(z) dz*dz

    is proportional to f(z0) much in the same way that

    (1/2π)∫ eiy(x - x0)f(x) dxdy
    = ∫ δ(x - x0)f(x) dx
    = f(x0)


    Is this true? Could someone help convince me of it, or point me to a text?

    I would say that even if true, it would be incorrect to say that

    ∫ ez*(z - z0)dz* = δ(z - z0)

    because the integration over dz and dz* cannot be done independently in the same way that a surface integral over dxdy in the plane can (sometimes) be separated into independent integrations over x and y. Or can it?
     
  2. jcsd
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