- Show that \(\mathrm { i f } \ | a | < 1\) , then
\[ \int_ {0} ^ {2 \pi} \log | 1 - a e ^ {i \theta} | d \theta = 0. \]
Then, prove that the above result remains true if we assume only that \(| a | \le 1\)
Solution
First suppose \(|a|<1\). The power series \[ \Log(1-w)=-\sum_{n\ge1}\frac{w^n}{n},\qquad |w|<1, \] converges uniformly on \(|w|\le |a|\). Hence \[ \log|1-ae^{i\theta}|=\Re\Log(1-ae^{i\theta}) =-\Re\sum_{n\ge1}\frac{a^n e^{in\theta}}n. \] Termwise integration is justified by uniform convergence, and every exponential term has integral \(0\) over \([0,2\pi]\). Thus the integral is \(0\).
Now let \(|a|=1\). Write \(a=e^{i\phi}\). By translation of \(\theta\) it is enough to treat \(a=1\). Since \[ |1-e^{i\theta}|=2\left|\sin\frac\theta2\right|, \] we need to show \[ \int_0^{2\pi}\log\!\left(2\left|\sin\frac\theta2\right|\right)d\theta=0. \] For \(0<r<1\), the already-proved case gives \[ \int_0^{2\pi}\log|1-re^{i\theta}|\,d\theta=0. \] As \(r\uparrow1\), these functions converge pointwise off \(\theta=0,2\pi\) to \(\log\abs{1-e^{i\theta}}\). For \(\frac12\le r<1\), \[ |1-re^{i\theta}|^2=(1-r)^2+4r\sin^2\frac\theta2, \] which lies between \(2\sin^2\frac\theta2\) and \(4\). Hence \(\abs{\log|1-re^{i\theta}|}\le\log2+\abs{\log\bigl(\sqrt2\,\abs{\sin\frac\theta2}\bigr)}\), an integrable function of \(\theta\) on \([0,2\pi]\). By dominated convergence, the integrals converge to the boundary integral, which is therefore also \(0\).