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

A conformal map at a point $z_0$ is a map that preserves oriented angles between regular differentiable curves through $z_0$. In the differential formulation, it is real differentiable at $z_0$ and its derivative is an orientation-preserving similarity: a rotation followed by multiplication of lengths by a positive constant. The map is conformal on $U$ when it has this property at every point:Here $R_{\alpha }$ is a planar rotation. Conformality is a local property and does not itself require global injectivity. We use the oriented convention standard in complex analysis; a reflection reverses oriented angles.

At $z_0$, complex differentiability givesWriting $f^{\prime }(z_0)=\lambda e^{i\alpha }$, the assumption $f^{\prime }(z_0)\ne 0$ gives $\lambda >0$. Its real derivative is therefore the similarity $h\mapsto \lambda e^{i\alpha }h$, a rotation through $\alpha$ followed by scaling by $\lambda$.

More explicitly, for two differentiable curves ${\gamma }_1,{\gamma }_2$ with ${\gamma }_j(0)=z_0$ and nonzero tangent vectors, the chain rule givesTaking the complex argument of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus

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.

Suppose $F:\mathbb{D}\to \mathbb{C}$ were a holomorphic bijection. By the allowed nonvanishing derivative result, $F^{\prime }(z)\ne 0$ throughout $\mathbb{D}$. The holomorphic inverse function theorem gives a holomorphic inverse locally at every image point. Since $F$ is globally bijective, these local inverses agree with its global inverse $H:\mathbb{C}\to \mathbb{D}$.

Thus $H$ is an entire function with $|H|<1$. The Liouville theorem forces $H$ to be constant, contradicting its being an inverse. ThereforeThis is the analytic obstruction behind the fact that the complex plane and unit disc are homeomorphic but not conformally equivalent.

Yes. An explicit holomorphic surjection is the rational functionIts denominator never vanishes inside the unit disc. To prove surjectivity, take any $w\in \mathbb{C}$ and choose a square root $r$ of $w$ with $\mathrm{Re}r\ge 0$; one of the two opposite roots always has this property. Define $z=r/(1+r)$. Then $1+r\ne 0$ andso $|z|<1$. Also $z/(1-z)=r$, and consequently $F(z)=r^2=w$.

This is a holomorphic surjection from the unit disc onto the complex plane. It does not contradict part (i): it is not injective. For example, $F(i/(1+i))=F(-i/(1-i))=-1$, with two distinct arguments of modulus $1/\sqrt{2}$.