log|z| has no global harmonic conjugate on C{0}

Suppose $v$ were a harmonic conjugate of $\log \mathrm{\mid }z\mathrm{\mid }$ on $\mathbb{C}\setminus \{0\}$. Then $F(z):=\log \mathrm{\mid }z\mathrm{\mid }+iv(z)$ would be holomorphic there, and its exponential would satisfy $\exp (F(z))=\mathrm{\mid }z\mathrm{\mid }(\cos v(z)+i\sin v(z))\mathrm{.}$

The function $G(z):=\exp (F(z))\mathrm{/}z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ by [L3], and $\mathrm{\mid }G(z)\mathrm{\mid }=\mathrm{\mid }\exp (F(z))\mathrm{\mid }\mathrm{/}\mathrm{\mid }z\mathrm{\mid }=e^{\mathrm{Re}F(z)}\mathrm{/}\mathrm{\mid }z\mathrm{\mid }=1$ by step 1.1. If $G$ were nonconstant, [L4] would make its image open in $\mathbb{C}$, impossible because $G(\mathbb{C}\setminus \{0\})\subseteq \{\mathrm{\mid }w\mathrm{\mid }=1\}$. Hence $G$ is constant on $\mathbb{C}\setminus \{0\}$.

Since $G$ is constant, for every $z\mathrm{\ne }0$ one has $\exp (F(z)-F(1))=\frac{\exp (F(z))}{\exp (F(1))}=\frac{zG(z)}{G(1)}=z\mathrm{.}$ Thus $L(z):=F(z)-F(1)$ is a continuous logarithm on $\mathbb{C}\setminus \{0\}$, contradicting [L2].

Therefore $\log \mathrm{\mid }z\mathrm{\mid }$ has no global harmonic conjugate on $\mathbb{C}\setminus \{0\}$.