Harmonic conjugate of $\ln|z|$ in $\mathbb C \setminus \{0\}$

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 $u$ ?

Solution 1:

Suppose a harmonic function $u \colon \mathbb{C}\setminus \{0\} \to \mathbb{R}$ has a (global) harmonic conjugate $v$. Then $f \colon \mathbb{C}\setminus \{0\} \to \mathbb{C}$ given by $f(z) = u(z) + iv(z)$ is holomorphic, and therefore

$$\int_{\lvert z\rvert = 1} f'(z)\,dz = 0.$$

Now, for holomorphic $f = u + i v$ we have

$$f'(z) = \frac{\partial f}{\partial x}(z) = \frac{\partial u}{\partial x}(z) + i\frac{\partial v}{\partial x},$$

and with the Cauchy-Riemann equations, that becomes

$$f'(z) = \frac{\partial u}{\partial x}(z) - i \frac{\partial u}{\partial y}(z).$$

Now, for $u(z) = \ln \lvert z\rvert$, we find

$$\frac{\partial u}{\partial x}(x+iy) = \frac{x}{x^2+y^2}\quad\text{ and }\quad \frac{\partial u}{\partial y}(x+iy) = \frac{y}{x^2+y^2},$$

and thus

$$f'(z)\,dz = \frac{\overline{z}\,dz}{\lvert z\rvert^2} = \frac{dz}{z},$$

and it is well-known that

$$\int_{\lvert z\rvert = 1} \frac{dz}{z} = 2\pi i \neq 0.$$

Solution 2:

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.