The real part of a holomorphic function is harmonic

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be holomorphic on $U$. Let $u = \operatorname{Re} f$ and $v = \operatorname{Im} f$, as real functions of $(x, y)$ on $U$, with $z = x + iy$. Then $u$ and $v$ are twice continuously differentiable, and $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \qquad \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$$ on $U$. Thus $u$ and $v$ are harmonic functions on $U$.

Remarks

Proof: $f$ is infinitely differentiable. By the Cauchy–Riemann equations, $u_{xx} = v_{yx} = v_{xy} = -u_{yy}$, and the same computation works for $v$. For example, $\operatorname{Re}(z^2) = x^2 - y^2$ is harmonic, and so is $\operatorname{Re}(e^z) = e^x \cos y$. The function $v$ is a harmonic conjugate of $u$.

Explanations

L1 human

A harmonic function is a function of two real variables whose two pure second derivatives add up to $0$. Each holomorphic function gives two harmonic functions for free: its real part and its imaginary part.