Real and Imaginary Parts of Holomorphic Function are Harmonic

Theorem
Let $f: U \to \C$ be a holomorphic complex function on $U \subseteq \C$.

Let $u, v: D \to \C$ be real-valued functions defined as:
 * $\map u {x, y} = \map \Re {\map f {x + i y} }$
 * $\map v {x, y} = \map \Im {\map f {x + i y} }$

where $D = \set {\tuple {x, y}: x + i y \in U}$

Then, $u$ and $v$ are harmonic functions.

Proof
By Cauchy-Riemann Equations, $u$ and $v$ satisfy:
 * $\dfrac {\partial u} {\partial x} = \dfrac {\partial v} {\partial y}$
 * $\dfrac {\partial u} {\partial y} = -\dfrac {\partial v} {\partial x}$

Consider the $x$ partial of the first equation, and the $y$ partial of the second:
 * $\dfrac {\partial^2 u} {\partial x^2} = \dfrac {\partial^2 v} {\partial x \partial y}$
 * $\dfrac {\partial^2 u} {\partial y^2} = -\dfrac {\partial^2 v} {\partial y \partial x}$.

By Clairaut's Theorem, $\dfrac {\partial^2 v} {\partial x \partial y} = \dfrac {\partial^2 v} {\partial y \partial x}$.

Thus, the sum of the two equations yields:
 * $\dfrac {\partial^2 u} {\partial x^2} + \dfrac {\partial^2 u} {\partial y^2} = 0$

So $u$ is a harmonic function by definition.

Now, consider the $y$ partial of the first equation, and the $x$ partial of the second:
 * $\dfrac {\partial^2 u} {\partial x \partial y} = \dfrac {\partial^2 v} {\partial y^2}$
 * $\dfrac {\partial^2 u} {\partial y \partial x} = -\dfrac {\partial^2 v} {\partial x^2}$

By Clairaut's Theorem, $\dfrac {\partial^2 u} {\partial x \partial y} = \dfrac {\partial^2 u} {\partial y \partial x}$

Thus, the difference of the two equations yields:
 * $0 = \dfrac {\partial^2 v} {\partial y^2} + \dfrac {\partial^2 v} {\partial x^2}$

So $v$ is a harmonic function by definition.