If $\psi_\alpha(z) = (\alpha-z)/(1-\bar{\alpha}z)$ for $|\alpha|<1$, prove that $${1\over \pi}\iint_{\Bbb D}|\psi'_\alpha|^2\ dxdy = 1.$$ [Hint: The integral can be evaluated without a calculation.]
Direct calculation shows that $\psi'_\alpha(z) = (|\alpha|^2-1)/(1-\bar{\alpha}z)^2$. Note that $|\psi_\alpha'(z)|^2 = \psi'_\alpha(z)\overline{\psi'_\alpha(z)}$. Maybe I can consider the mean value property of harmonic function (because there's a conjugate). But the product of harmonic function is generally not harmonic and I'm stuck at this point. Please help.
Use the multivariate change of coordinates formula, with $\psi_\alpha$ as your change of coordinates. Note that $\psi_\alpha$ is a holomorphic bijection of the disk to itself!