Use the Wirtinger-operator identity for the Laplacian to prove that if \(f\) is holomorphic in an open set \(\Omega\), then the real and imaginary parts of \(f\) are harmonic; that is, their Laplacians vanish.
Solution
1. Write \(f=u+iv\), where \(u,v:\Omega\to\mathbb R\).
Every complex-valued function has uniquely determined real and imaginary parts.
2. Since \(f\) is holomorphic,\[ \frac{\partial f}{\partial\overline z}=0. \]
For a holomorphic function, the Cauchy-Riemann equations are equivalent to vanishing of the \(\overline z\)-Wirtinger derivative.
3. Hence \(\Delta f=0\).
Holomorphic functions are smooth, so the identity \(4\partial_z\partial_{\overline z}=\Delta\) of E-SS1.EX-10 applies. Thus \[ \Delta f =4\frac{\partial}{\partial z}\frac{\partial f}{\partial\overline z} =4\frac{\partial}{\partial z}(0) =0. \]
4. One has\[ \Delta f=\Delta u+i\Delta v. \]
The Laplacian is real-linear and acts componentwise on \(f=u+iv\).
5. Therefore \(\Delta u=0\) and \(\Delta v=0\) on \(\Omega\).
By 3 and 4, \[ \Delta u+i\Delta v=0. \] Both \(\Delta u\) and \(\Delta v\) are real-valued, so the real and imaginary parts vanish separately. Thus \(u\) and \(v\) are harmonic.