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The integral of a harmonic function

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that:

    [tex]\frac{1}{2\pi}\int_0^{2\pi}log|re^{i\theta} - z_0|d\theta = \left\{\begin{matrix}
    log|z_0| & if & |z_0| < r \\
    log|r| & if & |z_0| > r

    2. Relevant equations

    The function log|z| is harmonic in the slit plane since it is the real part of log(z) which is analytic in the slit plane. I'm pretty sure I'm supposed to use the fact that it's harmonic.

    Harmonic functions have the Mean Value Property which states that the average of a harmonic function on the boundary of a ball is equal to the value of the harmonic function at the center point of this ball.

    3. The attempt at a solution

    What I don't understand is that this integral is the average of log|z| along a circle of radius r centered at [itex]z_0[/itex], and thus by the Mean Value Property I feel that it should always equal [itex]log|z_0|[/itex] for [itex]log|z| \neq r[/itex]. I also feel like we should have the requirement that [itex]z_0 \neq 0[/itex] since the logarithm of zero is negative infinity. What am I missing here? Thanks.
  2. jcsd
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