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Complex Analysis Question

  1. Apr 7, 2006 #1
    Consider the function:

    g(z(t)) = i*f '(c+it)/(f(c+it) - a)

    Where {-d <= t < d}

    If we let z = c+it

    By change of variables don't we get:

    Line integral of g(z(t)) = i ln[f(c+it) - a]

    evaluated from t = - d to t = d?

    note: ln is the natural log.

    Inquisitively,

    Edwin G. Schasteen
     
  2. jcsd
  3. Apr 8, 2006 #2
    You may consider the line integral as the complex path integral

    [tex]\int_\gamma\, \frac{f'(z)}{f(z)-a}\,dz[/tex]

    where [tex]\gamma(t) = c + it[/tex], for [tex]-d\le t\le d[/tex].

    Since the integrand has the trivial primitive [tex]G(z) = \ln(f(z)-a)[/tex], you may indeed consider that:

    [tex]\int_\gamma\, \frac{f'(z)}{f(z)-a}\,dz \quad = \quad G(c+id) - G(c-id)[/tex]

    You should be cautious anyway, since for this integral to be right, it must be [tex]f(z) \ne a[/tex] along [tex]\gamma[/tex]. Assuming that g(z) has no singularities in a domain containing the path [tex]\gamma[/tex], I think you can safely consider the primitive [tex]G(z)[/tex] as correct to compute the integral.
     
  4. Apr 8, 2006 #3
    That makes sense.

    Thanks! I appreciate your help!

    Best Regards,

    Edwin
     
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