Let \(G = \mathbb{C} \setminus \{0\}\), and let \(h : G \to \mathbb{R}\) be harmonic.
Show that \(g = h_x - ih_y\) is a holomorphic function in \(G\), and that its residue at \(0\) is a real number.
Show that there exist a constant \(c\) and a holomorphic function \(f : G \to \mathbb{C}\) such that \[ h(z) = c\log|z| + \operatorname{Re} f(z) \quad \forall z \in G. \]
Solution
Since \(h\) is harmonic, \[ g=h_x-ih_y=2\frac{\partial h}{\partial z} \] is holomorphic on \(G\). For a positively oriented circle \(|z|=r\), parametrized by \(z=re^{it}\), \[ \int_{|z|=r}g(z)\,dz =\int_0^{2\pi}(h_x-ih_y)(-r\sin t+ir\cos t)\,dt. \] Its real part is the integral of the tangential derivative of \(h\) around the circle, hence is \(0\). Therefore the integral is purely imaginary, and \[ \operatorname{Res}_0 g=\frac1{2\pi i}\int_{|z|=r}g(z)\,dz \] is real. Write this residue as \(c\in\mathbb R\).
Then \(g(z)-c/z\) has zero period around a generator of \(\mathbb C\setminus\{0\}\), hence all its periods vanish. It therefore has a holomorphic primitive \(F\) on \(G\). Since \[ 2\frac{\partial}{\partial z}\bigl(h-c\log|z|-\operatorname{Re}F\bigr)=0, \] the real-valued function in parentheses is constant. Absorb that real constant into \(F\). We obtain \[ \boxed{h(z)=c\log|z|+\operatorname{Re}F(z)}. \]