 Solution

Let $V=f(U)$, the right half-disc from part (i). Write $w=r{e}^{it}$ with $0<r<1$ and $-\pi /2<t<\pi /2$. Squaring doubles the angle and squares the radius, giving a holomorphic bijectionIts 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 $s=-t$, $0\le t<1$, it givesConsequently $T((-1,0])=\{iy:y\ge 1\}$. Since $T$ is a bijection on the whole disc, deleting this interval deletes exactly that ray in the image.

The required function composition is thereforeEach component is a conformal bijection between the indicated domains. The only possible denominator zeros of the simplified expression are $z=\pm 1$, outside $U$, andsince $i$ 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.