= Solution
For $z=x+iy\in\mathbb H$, direct calculation gives
$$
|f(z)|^2=\frac{x^2+(y-1)^2}{x^2+(y+1)^2}<1.
$$
Conversely, solving $w=(z-i)/(z+i)$ gives
$$
z=i\frac{1+w}{1-w},\qquad\operatorname{Im}z=\frac{1-|w|^2}{|1-w|^2}>0\quad(|w|<1).
$$
The denominator cannot vanish inside the <unit disc>, so these inverse formulas prove $\boxed{f(\mathbb H)=\mathbb D}$. This is the <Cayley transform between the half-plane and disk>.
For $z\in U$, we also require $|z|>1$. In inverse coordinates this becomes
$$
\frac{|1+w|}{|1-w|}>1\ \Longleftrightarrow\ |1+w|^2-|1-w|^2=4\operatorname{Re}w>0.
$$
The equivalence holds in both directions, so
$$
\boxed{f(U)=\{w\in\mathbb C:|w|<1,\ \operatorname{Re}w>0\}}.
$$
Thus the exterior of the unit semicircle maps to the right half of the <unit disc>. The circular arc maps to its imaginary diameter, which is excluded because the source inequality is strict.
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