Are holomorphic functions divergence free?

Question

I am attempting to understand a proof in a paper I'm reading. I am stuck on the first line. We construct a box, R, in the complex plane. Then, by Green's theorem, we have

$$\int_{\partial R}\frac{\partial u}{\partial n} = 0$$

All we know about u is that it's holomorphic on a region containing the box R. If we are using Green's theorem, it seems to me that we need to show the divergence of u is zero since we are computing the flux of u around a closed path. I thought maybe this would follow from the Cauchy-Riemann equations, but it doesn't, since we have $\partial_{x} Re(u) = \partial_{y} Im(u)$ but we would need $\partial_{x} Re(u) + \partial_{y} Im(u) = 0$.

I am not good with complex analysis, so any help would be greatly appreciated!

Answer 1

If we use the Gauss theorem, we can rewrite \begin{equation} \displaystyle \int_{\partial R} \frac{\partial u}{\partial n} = \displaystyle \int_{\partial R} \nabla u \cdot n \underbrace{=}_{\text{by Gauss}} \displaystyle \int_{R} \nabla^2 u = \displaystyle \int_{R} \nabla^2 \left[ Re(u) + i Im(u) \right] = 0 \end{equation} where we get the last equality from the fact that the real and imaginary parts of a holomorphic function are harmonic. You get this fact from the Cauchy-Riemann equations. If we let $Re(u) = h$ and $Im(u) = g$, we have (by C-R) \begin{align} \partial_{x} h &= \partial_{y} g \\ \partial_{y} h &= -\partial_{x} g \\ \end{align} Taking $\partial_{x}$ of the first equation and adding this to $\partial_{y}$ of the second will give us that $h$ is harmonic. While taking $\partial_{y}$ of the first equation and subtracting this from $\partial_{x}$ of the second will give us that $g$ is harmonic. (note that this is due to mixed partial derivatives commuting). Hope that this helps!