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Complex Integral

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume that f(z) is analytic and that f'(z) is continuous in a region that contains a closed curve [tex]\gamma[/tex]. Show that
    [tex] \int_\gamma \overline{f(z)} f'(z) dz[/tex]

    is purely imaginary.


    2. Relevant equations

    If f(z) is holomorphic on the region containing a closed curve [tex]\gamma[/tex] or if f(z) has a primitive (we have not yet established a relationship between these two properties), then

    [tex]\int_\gamma f(z)dz = 0.[/tex]

    And while we did not prove that analytic functions are holomorphic, that's not hard to verify if necessary.

    3. The attempt at a solution

    If we let [tex]f(z) = u(z) + i v(z)[/tex], where u and v are functions from the complex plane to the real line, then

    [tex]\int_\gamma \overline{f(z)} f'(z) dz = \int f(z) f'(z) dz - 2 i \int v(z) f'(z) dz = 0 - 2 i \int_\gamma v(z) f'(z)dz[/tex]

    So I have to verify that the latter integral is real valued, and this is where I am stuck. I could parametrize [tex]z = \gamma(t)[/tex] where [tex]t \in [0,1][/tex] but I am not sure that helps me in any way.

    [tex]- 2 i \int_\gamma v(z) f'(z)dz = -2 i \int_0^1 v(\gamma(t)) f'(\gamma(t)) \gamma'(t) dt = 2 i \int_0^1 v'(\gamma(t))f'(\gamma(t))\gamma'(t) \gamma'(t) dt[/tex]

    (The last step comes from integration by parts, with one term = 0 because gamma is a closed curve.)

    Is there any reason to believe that this last integral is real-valued?
     
    Last edited: Oct 4, 2009
  2. jcsd
  3. Oct 4, 2009 #2

    jbunniii

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    I am not sure what you did here.

    If [itex]f(z) = u(z) + iv(z)[/itex], then [tex]\overline{f(z)} = u(z) - i v(z)[/tex] and [itex]f'(z) = u'(z) + i v'(z)[/tex], so

    [tex]\begin{align*}\int_{\gamma} \overline{f(z)} f'(z) dz &= \int_{\gamma} (u(z) - i v(z)) (u'(z) + i v'(z)) dz \\
    &= \int_{\gamma} (u(z) u'(z) + v(z) v'(z)) dz +i \int_{\gamma} (u(z) v'(z) - v(z) u'(z)) dz \end{align*}[/tex]

    All of [itex]u(z)[/itex], [itex]u'(z)[/itex], [itex]v(z)[/itex], and [itex]v'(z)[/itex] are real, so if you can show that the first integral is zero then you're done.
     
  4. Oct 5, 2009 #3

    But [tex]dz[/tex] is not real. Does that mess with the value of the last integral?

    Thanks for your thoughts!
     
  5. Oct 5, 2009 #4

    Dick

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    You seem to be having a hard time finding a good expression for f'(z)*dz. You can do it several ways. (df/dz)*dz=df. If f(z)=u(z)+iv(z) and z=x+iy, that makes it (u+iv)_x*dx+(u+iv)_y*dy. (Underscores are partial derivatives). Or you could use f'(z)=(u+iv)_x or f'(z)=(1/i)*(u+iv)_y and dz=dx+idy. You can show they are all equal with Cauchy-Riemann. Now just collect the real and imaginary parts and figure out whether they are exact differentials. Take it from there.
     
    Last edited: Oct 5, 2009
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