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