$\frac1\pi\iint_{\mathbb{D}}|\psi_\alpha'|^2=1$ and $\frac1\pi\iint_{\mathbb{D}}|\psi_\alpha'|=\frac{1-|\alpha|^2}{|\alpha|^2}\log\frac1{1-|\alpha|^2}$ for $\psi_\alpha(z)=\frac{\alpha-z}{1-\bar\alpha z}$

Part 1: S&S claim this can be done without a calculation – maybe \(\abs{\psi_a'(z)}^2\) is constant on a circle of radius \(r\)…?

Part 2: \[ {1\over \pi} \iint_\DD \abs{\psi_a'(z)} \dz &= {1\over \pi} \iint_\DD \abs{1- \abs{a}^2 \over (1-\bar{a} z)^2 } \dz \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over \abs{ 1-\bar{a} re^{it} }^2 } r\dt\dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over (1-\bar a re^{it} )(1 - a re^{-it}) } r\dt\dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 \oint_{\bd \DD} {1 \over (1-\bar a rz )(1 - a r \bar z) } {r\over iz} \dz\dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 \oint_{\bd \DD} {1 \over (1-\bar a rz )(1 - a r \inverseof{z}) } {r\over iz} \dz\dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 {r\over i} \cdot \oint_{\bd \DD} {1 \over (1-\bar a rz )(z - a r) } \dz\dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i \Res_{z=ar} {1 \over (1-\bar a rz )(z - a r) } \dr \\ &= {1- \abs{a}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i {1 \over (1-\bar a rz ) }\evalfrom_{z=ar} \dr \\ &= {2\pi \qty{ 1- \abs{a}^2} \over \pi} \int_0^1 {r \over 1- \abs{a}^2r^2 } \dr \\ &= {1-\abs{a}^2 \over \abs{a}^2} \log\qty{1 \over 1 - \abs{a}^2 } .\]