Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 6 / b

For $z=x+iy\in \mathbb{H}$, direct calculation givesConversely, solving $w=(z-i)/(z+i)$ givesThe 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 becomesThe equivalence holds in both directions, soThus 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.

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.