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?