= Solution
Let $V=f(U)$, the right half-disc from part (i). Write $w=re^{it}$ with $0<r<1$ and $-\pi/2<t<\pi/2$. Squaring doubles the angle and squares the radius, giving a <holomorphic> bijection
$$
w\longmapsto w^2:\ V\longrightarrow W=\mathbb D\setminus(-1,0].
$$
Its inverse is the unique square root with positive real part, and its derivative $2w$ is nonzero on $V$. The cut includes zero, because $0\notin V$.
Next, the <Möbius transformation> $T(s)=i(1-s)/(1+s)$ is a <conformal> bijection from $\mathbb D$ onto $\mathbb H$: it is the inverse <Cayley transform between the half-plane and disk> composed with $s\mapsto-s$. For $s=-t$, $0\leq t<1$, it gives
$$
T(-t)=i\frac{1+t}{1-t}.
$$
Consequently $T((-1,0])=\{iy:y\geq1\}$. Since $T$ is a bijection on the whole disc, deleting this interval deletes exactly that ray in the image.
The required <function composition> is therefore
$$
\boxed{g(z)=i\frac{1-f(z)^2}{1+f(z)^2}=\frac{2z}{1-z^2}}.
$$
Each component is a <conformal> bijection between the indicated domains. The only possible denominator zeros of the simplified expression are $z=\pm1$, outside $U$, and
$$
g'(z)=\frac{2(1+z^2)}{(1-z^2)^2}\ne0\quad(z\in U),
$$
since $i$ is on the excluded semicircle and $-i$ is outside the <upper half-plane>. This also checks the local <conformality> of the <conformal map from the exterior half-disc to a slit half-plane>.
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