TITLE: Harmonic conjugate of $\ln|z|$ in $\mathbb C \setminus \{0\}$
QUESTION [7 upvotes]: How I prove that $u(z)=\ln |z|$ has no harmonic conjugate in $\mathbb C \setminus \{0\}$ ?

I found from C-R equation that $v(z)=\arctan(y/x)$ , which is defined in  $\mathbb C \setminus \{0\}$ . So why it is not a harmonic conjugate of $u$ ?

REPLY [7 votes]: Let $\log z = \ln|z| + i\text { arg }z$ denote the principal value logarithm, which is holomorphic on $U =\mathbb C \setminus (-\infty,0].$ Suppose $\ln |z| + iv(z)$ is holomorphic on $\mathbb C\setminus \{0\}.$ Then on $U,$ $\log z - (\ln |z| + iv(z))$ is holomorphic and purely imaginary. By the open mapping theorem, or the CR equations, this function is constant on $U.$ This implies $\ln |z| + iv(z)$  has the same jump discontinuity problem that $\log z$ has at every point on the negative real axis. It follows that $\ln |z| + iv(z)$ is not holomorphic on $\mathbb C\setminus \{0\},$ contradiction.