Show that:

$$\frac{1}{2\pi}\int_0^{2\pi}\log|re^{i\theta} - z_0|d\theta = \begin{cases} \log|z_0| & if & |z_0| < r \\ \log|r| & if & |z_0| > r \end{cases}.$$

I know $\log|z|$ is a harmonic function in the slit plane since it is the real part of the analytic function $\log(z)$ on the slit plane. What I don't understand is that for $|z_0| > r$ this integral is exactly the average of a harmonic function along the boundary of a disk upon which $\log|z|$ is harmonic, thus by the Mean-Value Property it should be equal to $\log|z_0|$, yet it's supposed to be $\log|r|$. What is going on?

  • $\begingroup$ The conditions must be reversed! If $|z_0|$ is enormous, $\log|r e^{i\theta}-z_0|$ should be large. If $z_0 = 0$, $\log|r e^{i\theta}-z_0| = \log |r|$. $\endgroup$ – Robert Israel Apr 21 '12 at 1:08
  • $\begingroup$ Can you expand on this? Are you saying that the problem as stated is incorrect, and that I should switch the inequalities? Because that would make a lot more sense. $\endgroup$ – Thoth Apr 21 '12 at 1:11
  • $\begingroup$ Yes, that's what I am saying. $\endgroup$ – Robert Israel Apr 22 '12 at 5:51

It's the average value of $\log|rz - z_0|$ over the unit circle $|z| = 1$. If $|z_0| > r$, the function is harmonic throughout the interior of the circle and is continuous on the boundary. So by the mean value theorem this average value is the value of the function at the center $z = 0$, namely $\log|z_0|$.

In the case where $|z_0| < r$, since $|re^{i\theta} - z_0| = |r - z_0e^{-i\theta}|$, your integral is the same as $${1 \over 2\pi}\int_0^{2\pi} \log|r - z_0e^{-i\theta}|\,d\theta$$ Changing variables from $\theta$ to $-\theta$, this equals $${1 \over 2\pi}\int_0^{2\pi} \log|r - z_0e^{i\theta}|\,d\theta$$ Now we have the average value of the function $\log|r - z_0z|$ over the unit circle, and since $|z_0| < r$ this function is harmonic inside the circle and continuous on the boundary. Thus the average value is that at the origin, namely $\log|r|$.

(Note as some commenters mentioned you have the values reversed.)

  • $\begingroup$ Great, thanks for the detailed response Zarrax! +1. $\endgroup$ – Thoth Apr 21 '12 at 1:48

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