- If \(\psi _ { \alpha } ( z ) = ( \alpha - z ) / ( 1 - \overline { { \alpha } } z )\) for \(| \alpha | < 1\) , prove that
\[ \frac {1}{\pi} \iint_ {\mathbb {D}} | \psi_ {\alpha} ^ {\prime} | ^ {2} d x d y = 1 \quad \mathrm{and} \quad \frac {1}{\pi} \iint_ {\mathbb {D}} | \psi_ {\alpha} ^ {\prime} | d x d y = \frac {1 - | \alpha | ^ {2}}{| \alpha | ^ {2}} \log \frac {1}{1 - | \alpha | ^ {2}}, \]
where in the case \(\alpha = 0\) the expression on the right is understood as the limit as \(| \alpha | \to 0\)
[Hint: The first integral can be evaluated without a calculation. For the second, use polar coordinates, and for each fixed r use contour integration to evaluate the integral in θ.]
Solution
Let \(r=|\alpha|\). Since \[ \psi_\alpha'(z) =-\frac{1-r^2}{(1-\overline\alpha z)^2}, \] and \(\psi_\alpha\) is an automorphism of \(\mathbb D\), the change-of-variables formula for area gives \[ \iint_{\mathbb D}|\psi_\alpha'(z)|^2\,dx\,dy =\operatorname{Area}(\psi_\alpha(\mathbb D)) =\pi. \] Therefore \[ \frac1\pi\iint_{\mathbb D}|\psi_\alpha'|^2\,dx\,dy=1. \]
For the second integral, rotate coordinates so that \(\alpha=r\ge0\). In polar coordinates \(z=\rho e^{i\theta}\), \[ \frac1\pi\iint_{\mathbb D}|\psi_\alpha'(z)|\,dx\,dy =\frac{1-r^2}{\pi} \int_0^1\int_0^{2\pi} \frac{\rho\,d\theta\,d\rho}{|1-r\rho e^{i\theta}|^2}. \] The Poisson-kernel identity gives, for \(0\le r\rho<1\), \[ \int_0^{2\pi}\frac{d\theta}{1-2r\rho\cos\theta+r^2\rho^2} =\frac{2\pi}{1-r^2\rho^2}. \] Hence \[ \frac1\pi\iint_{\mathbb D}|\psi_\alpha'| =2(1-r^2)\int_0^1\frac{\rho}{1-r^2\rho^2}\,d\rho =\frac{1-r^2}{r^2}\log\frac1{1-r^2}. \] As \(r\to0\), the right-hand side tends to \(1\), which agrees with \(\psi_0(z)=-z\) and \(|\psi_0'|=1\).