Let \(\psi_\alpha(z) = \frac{\alpha - z}{1 - \bar{\alpha}z}\) with \(\abs{\alpha} < 1\) and \(\mathbb{D} = \{z : \abs{z} < 1\}\). Prove that
\(\displaystyle\frac{1}{\pi}\iint_{\mathbb{D}} \abs{\psi_\alpha'}^2\,dx\,dy = 1\).
\(\displaystyle\frac{1}{\pi}\iint_{\mathbb{D}} \abs{\psi_\alpha'}\,dx\,dy = \frac{1 - \abs{\alpha}^2}{\abs{\alpha}^2}\log\frac{1}{1 - \abs{\alpha}^2}\).
The source prints this problem twice, as questions 24 and 30 of Spring 2015. Under question 30 it also prints “Prove that \(f(z) = -\frac{1}{2}\left(z + \frac{1}{z}\right)\) is a conformal map from half disc \(\{z = x + iy : \abs{z} < 1,\ y > 0\}\) to upper half plane \(\mathbb{H} = \{z = x + iy : y > 0\}\)”, which is question 25 of the same list. It is a separate problem, carried by its own card.