6.4: A second Proof that u and v are Harmonic

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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.