Real and Imaginary Parts of Holomorphic Function are Harmonic

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Theorem

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.

Let $f: D \to \C$ be a holomorphic complex function on $D$.

Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined as:

\(\ds \map u {x, y}\) \(=\) \(\ds \map \Re {\map f {x + i y} }\)
\(\ds \map v {x, y}\) \(=\) \(\ds \map \Im {\map f {x + i y} }\)


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


Proof

By Cauchy-Riemann Equations, $u$ and $v$ satisfy:

\(\text {(1)}: \quad\) \(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \dfrac {\partial v} {\partial y}\)
\(\text {(2)}: \quad\) \(\ds \dfrac {\partial u} {\partial y}\) \(=\) \(\ds -\dfrac {\partial v} {\partial x}\)



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Consider the partial derivative of $(1)$ with respect to $x$, and the partial derivative of $(2)$ with respect to $y$:

\(\text {(3)}: \quad\) \(\ds \dfrac {\partial^2 u} {\partial x^2}\) \(=\) \(\ds \dfrac {\partial^2 v} {\partial x \partial y}\)
\(\text {(4)}: \quad\) \(\ds \dfrac {\partial^2 u} {\partial y^2}\) \(=\) \(\ds -\dfrac {\partial^2 v} {\partial y \partial x}\)

By Clairaut's Theorem:

$\dfrac {\partial^2 v} {\partial x \partial y} = \dfrac {\partial^2 v} {\partial y \partial x}$


Thus $(3) + (4)$ yields:

$\dfrac {\partial^2 u} {\partial x^2} + \dfrac {\partial^2 u} {\partial y^2} = 0$

So $u$ is a harmonic function by definition.

$\Box$


Now consider the partial derivative of $(1)$ with respect to $y$, and the partial derivative of $(2)$ with respect to $x$:

\(\text {(5)}: \quad\) \(\ds \dfrac {\partial^2 u} {\partial x \partial y}\) \(=\) \(\ds \dfrac {\partial^2 v} {\partial y^2}\)
\(\text {(6)}: \quad\) \(\ds \dfrac {\partial^2 u} {\partial y \partial x}\) \(=\) \(\ds -\dfrac {\partial^2 v} {\partial x^2}\)

By Clairaut's Theorem :

$\dfrac {\partial^2 u} {\partial x \partial y} = \dfrac {\partial^2 u} {\partial y \partial x}$

Thus $(5) - (6)$ yields:

$0 = \dfrac {\partial^2 v} {\partial y^2} + \dfrac {\partial^2 v} {\partial x^2}$

So $v$ is a harmonic function by definition.

$\blacksquare$


Sources

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