Prove that the map \[ w=\frac{z-1}{z+1} \] maps the first quadrant in the plane conformally onto the upper open unit disk \[ \Delta^+=\{w=u+iv:\ u^2+v^2<1\text{ and }v>0\}. \]
- Is the intersection of \(\Delta^+\) with the first quadrant conformally equivalent to the upper half-plane \(\{z=x+iy:\ y>0\}\)?
Solution
Let \[ T(z)=\frac{z-1}{z+1}. \] For \(z=x+iy\) with \(x>0\) and \(y>0\), \[ |T(z)|^2=\frac{(x-1)^2+y^2}{(x+1)^2+y^2}<1 \] because \(x>0\), while \[ \operatorname{Im}T(z)=\frac{2y}{|z+1|^2}>0. \] Thus the first quadrant maps into \(\Delta^+\).
Conversely, \[ T^{-1}(w)=\frac{1+w}{1-w}. \] If \(|w|<1\), then \[ \operatorname{Re}T^{-1}(w)=\frac{1-|w|^2}{|1-w|^2}>0, \] and if \(\operatorname{Im}w>0\), then \[ \operatorname{Im}T^{-1}(w)=\frac{2\operatorname{Im}w}{|1-w|^2}>0. \] Hence \(T^{-1}\) maps \(\Delta^+\) into the first quadrant. Therefore \(T\) is a conformal bijection from the first quadrant onto \(\Delta^+\).
For (a), the intersection of \(\Delta^+\) with the first quadrant is the quarter disk \[ Q=\{w:|w|<1,\ 0<\arg w<\pi/2\}. \] It is a nonempty simply connected proper domain, so by the Riemann mapping theorem it is conformally equivalent to \(\mathbb D\), and hence to the upper half-plane via a Cayley transform. Thus the answer is yes.