how can it be proved that $\text{Log}|z-z_0|$ is harmonic in $\mathbb{C}\setminus\{z_0\}$?

Question

I know it can be proved in almost all of $\mathbb{C}\setminus\{z_0\}$, (exept the straight line from $z_0$ parallel to $[-\infty,0]$), as the real part of the holomorphic function $\text{Log}(z-z_0)$.

Answer 1

• Any holomorphic branch of $\log(z-z_0)$ in some domain $D$ satisfies $$ \begin{align} &z-z_0 = e^{\log(z-z_0)} \\ \implies& |z-z_0| = e^{\operatorname{Re} \log(z-z_0)} \\ \implies& \log|z-z_0| = \operatorname{Re} \bigl(\log(z-z_0)\bigr) \end{align} $$ in $D$, not only the principal part.

• Every point in $\mathbb{C}\setminus\{z_0\}$ has a neighborhood (e.g. a disk) in which a holomorphic branch of $\log(z-z_0)$ exists.

Combining these two facts we can conclude that $\log|z-z_0|$ is harmonic everywhere in $\mathbb{C}\setminus\{z_0\}$.

Answer 2

Can also just calculate the Laplacian directly using $\ln |z| = \frac{1}{2} \ln (x^2+y^2)$. Or use the complex form of the Laplacian, $\frac 1 4 \frac{d^2}{dzd\bar{z}}$ on $\ln |z| = \frac 1 2 \ln z \bar{z} = \frac 1 2 \ln z + \frac 1 2 \ln \bar{z}$.