Re(f) and Im(f) of an Analytic Function f are Harmonic Functions

# The Real and Imaginary Parts of an Analytic Function are Harmonic Functions

Recall from the Harmonic Functions page that if $A \subseteq \mathbb{R}^2$ is open, $u : A \to \mathbb{R}$, and the unmixed second partial derivatives of $u$ exist then the Laplacian of $u$ is defined as:

(1)\begin{align} \quad \Delta (u) = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \end{align}

Moreover, we said that $u$ is harmonic on $A$ if the Laplacian $\Delta (u)$ equals to $0$ on all of $A$.

We will now look at a remarkable theorem which says that if $f = u + iv$ is analytic on $A$ then both $u$ and $v$ must be harmonic on $A$.

Theorem 1: Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ where $f = u + iv$. If $f$ is analytic on $A$ then $u$ and $v$ are harmonic on $A$. |

**Proof:**Suppose that $f$ is analytic on $A$. Then by the Cauchy-Riemann theorem we have that $f$ (amongst other things) satisfies the Cauchy-Riemann equations on all of $A$, i.e.:

\begin{align} \quad \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad (1) \end{align}

(3)
\begin{align} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \quad (2) \end{align}

- We differentiate $(1)$ above with respect to $x$ to get:

\begin{align} \quad \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 v}{\partial x \partial y} \end{align}

- We differentiate the $(2)$ above with respect to $y$ to get:

\begin{align} \quad \frac{\partial^2 u}{\partial y^2} = -\frac{\partial^2 v}{\partial y \partial x} \end{align}

- The second order mixed partial derivatives of $u$ are equal because they are continuous (Clairaut's Theorem), and so we have that:

\begin{align} \quad \frac{\partial^2 u}{\partial x^2} = - \frac{\partial^2 u}{\partial y^2} \end{align}

- Therefore $\displaystyle{\Delta (u) = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0}$ on all of $A$, so $u$ is harmonic on $A$.

- Similarly, we differentiate $(1)$ above with respect to $y$ to get:

\begin{align} \quad \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 v}{\partial y^2} \end{align}

- And we differentiate $(2)$ above with respect to $x$ to get:

\begin{align} \quad \frac{\partial ^2 u}{\partial x \partial y} = -\frac{\partial^2 v}{\partial x^2} \end{align}

- Once again, the second order mixed partial derivatives of $v$ are equal since they are continuous and so we have that:

\begin{align} \quad \frac{\partial^2 v}{\partial x^2} = -\frac{\partial^2 v}{\partial y^2} \end{align}

- Therefore $\displaystyle{\Delta (v) = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0}$ on all of $A$, so $v$ is also harmonic on $A$. $\blacksquare$