6.4: A second Proof that u and v are Harmonic

This fact that $u$ and $v$ are harmonic is important enough that we will give a second proof using Cauchy’s integral formula. One benefit of this proof is that it reminds us that Cauchy’s integral formula can transfer a general question on analytic functions to a question about the function $1/z$. We start with an easy to derive fact.

Fact

The real and imaginary parts of $f(z) = 1/z$ are harmonic away from the origin. Likewise for

$$g(z) = f(z - a) = \dfrac{1}{z - a} \nonumber$$

away from the point $z = a$.

Proof

We have

$$\dfrac{1}{z} = \dfrac{x}{x^2 + y^2} - i \dfrac{y}{x^2 + y^2}. \nonumber$$

It is a simple matter to apply the Laplacian and see that you get 0. We’ll leave the algebra to you! The statement about $g(z)$ follows in either exactly the same way, or by noting that the Laplacian is translation invariant.

Second proof that $f$ analytic implies $u$ and $v$ are harmonic. We are proving that if $f = u + iv$ is analytic then $u$ and $v$ are harmonic. So, suppose $f$ is analytic at the point $z_0$. This means there is a disk of some radius, say $r$, around $z_0$ where $f$ is analytic. Cauchy's formula says

$$f(z) = \dfrac{1}{2\pi i} \int_{C_r} \dfrac{f(w)}{w - z}\ dw, \nonumber$$

where $C_r$ is the circle $|w - z_0| = r$ and $z$ is in the disk $|z - z_0| < r$.

Now, since the real and imaginary parts of $1/(w - z)$ are harmonic, the same must be true of the integral, which is limit of linear combinations of such functions. Since the circle is finite and $f$ is continuous, interchanging the order of integration and differentiation is not a problem.