Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2005 / 6 / ii / Solution

Use the Cayley transformFor $\mathrm{Im}z>0$, its denominator is nonzero, and direct subtraction gives $1-|w{|}^2=4\mathrm{Im}z/|z+i{|}^2>0$. Thus the image lies in the open unit disk. Conversely, for $|w|<1$ the inverse is $z=i(1+w)/(1-w)$, with $\mathrm{Im}z=(1-|w{|}^2)/|1-w{|}^2>0$. Hence it maps the upper half-plane bijectively onto the disk.