Find a conformal map from \(\DD\) to \(\HH\).
Solution
The inverse of the standard Cayley map \(w\mapsto(w-i)/(w+i)\) is \[ T(z)=i{1+z\over1-z}. \] For \(z\in\DD\), \[ \Im T(z) =\Re {1+z\over1-z} ={1-|z|^2\over |1-z|^2}>0, \] so \(T(\DD)\subseteq\HH\). Its inverse is \[ T^{-1}(w)={w-i\over w+i}, \] which maps \(\HH\) into \(\DD\). Thus \(T:\DD\to\HH\) is biholomorphic, hence conformal.