Let \(f=u+iv\) be holomorphic on \(|z|<1\), with \(u,v\) real-valued. Assume \[ u(0)^2=v(0)^2. \] Show that for every \(0<r<1\), \[ \int_0^{2\pi}u(re^{i\theta})^2\,d\theta = \int_0^{2\pi}v(re^{i\theta})^2\,d\theta. \]
Solution
1. The function\[ U(z)\coloneqq u(z)^2-v(z)^2 \]is harmonic on the unit disk.
Since \[ f=u+iv \] is holomorphic, so is \[ f^2 =(u^2-v^2)+2iuv. \] The real part of a holomorphic function is harmonic. Hence \[ U=\Re(f^2) \] is harmonic.
2. For every \(0<r<1\),\[ \frac1{2\pi} \int_0^{2\pi} \left( u(re^{i\theta})^2-v(re^{i\theta})^2 \right) \,d\theta = u(0)^2-v(0)^2. \]
Apply the mean-value property to the harmonic function \(U\) from step 1 on the circle of radius \(r\) centered at the origin.
3. For every \(0<r<1\),\[ \int_0^{2\pi} u(re^{i\theta})^2\,d\theta = \int_0^{2\pi} v(re^{i\theta})^2\,d\theta. \]
The hypothesis gives \[ u(0)^2-v(0)^2=0. \] Therefore step 2 gives \[ \int_0^{2\pi} \left( u(re^{i\theta})^2-v(re^{i\theta})^2 \right) \,d\theta =0. \] Move the second integral to the other side.
4. Q.E.D.
Step 3 is the required identity.