Proving that $f$ is analytic if $f$ and $z f(z)$ are harmonic

Question

If $f$ is harmonic and $zf(z)$ is harmonic, then $f$ is analytic. Please help me prove this. Thanks.

Answer 1

The idea is to show $f$ satisfies the Cauchy-Riemann equation. We will use the Wirtinger derivatives $$ \frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\qquad \frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right). $$ To begin with, observe that for $C^2$ functions, $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ commute, so the Laplacian $\Delta$ is then given by $$\Delta=4 \frac{\partial^2}{\partial \overline{z}\partial z}.$$

Denote $g(z):=zf(z)$. Since $f$ is assumed to be harmonic, it is in particular $C^2$, like $g$.

First, we compute $$ \frac{\partial g}{\partial z}=f+z\frac{\partial f }{\partial z} $$ whence $$ \frac{\partial^2 g}{\partial \overline{z}\partial z}=\frac{\partial f }{\partial \overline{z}}+\frac{\partial z }{\partial \overline{z}}\frac{\partial f }{\partial z}+z\frac{\partial^2f}{\partial \overline{z}\partial z}=\frac{\partial f }{\partial \overline{z}}+z\frac{\partial^2f}{\partial \overline{z}\partial z}. $$ We've only used the product rule for $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \overline{z}}$, and the fact that $\frac{\partial z}{\partial \overline{z}}=0$, which is easy to check. So $$ \frac{\partial f }{\partial \overline{z}}=\frac{1}{4}(\Delta g-z\Delta f). $$ It follows that if $f$ and $g$ are both harmonic, that is $\Delta f=\Delta g=0$, then $f$ satisfies the Cauchy-Riemann equation: $$ \frac{\partial f }{\partial \overline{z}}=0\qquad\iff\qquad \frac{\partial f }{\partial x}+i\frac{\partial f }{\partial y}=0. $$ As you can see here, a real differentiable function on $\mathbb{C}$ is holomorphic if and only if it satisfies the Cauchy-Riemann equation. So $f$ is holomorphic.