Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 6

A conformal map preserves oriented angles between regular smooth curves through each point. For a holomorphic map with nonzero derivative, the derivative also describes the local scale and rotation.

Indeed, if two curves ${\gamma }_1,{\gamma }_2$ meet at $z$ and have nonzero tangent vectors $v_1,v_2$, the real chain rule and complex differentiability giveBecause $f^{\prime }(z)\ne 0$, the image tangents are nonzero andTaking arguments proves preservation of oriented angles. Multiplication by $f^{\prime }(z)$ rotates every tangent by $\arg f^{\prime }(z)$ and scales every length by $|f^{\prime }(z)|$. Thus the hypothesis gives conformality at every point of $U$; global injectivity is not needed for this local conclusion.

A Möbius transformation maps generalized circles to generalized circles on the Riemann sphere. A generalized circle is an ordinary circle or a straight line together with the point at infinity. Its image is a straight line precisely when it contains infinity, equivalently when the preimage of infinity lies on the extended real axis.

If $c=0$, the pole is infinity itself and the transformation is the affine map $(a/d)z+b/d$, taking the extended real axis to a line. If $c\ne 0$, the pole is $-d/c$, which is real exactly when $d/c$ is real. Thus the full condition isEquivalently, it is $\mathrm{Im}(d\bar{c})=0$, a condition also valid when $c=0$. If it fails, the image misses infinity and is therefore an ordinary circle. The nonzero determinant ensures that none of these generalized circles collapses to a point.

Liouville's theorem states that every bounded entire function is constant:For example, the Cauchy derivative estimate on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ gives $f^{\prime }=0$ everywhere, proving the conclusion.

It suffices to construct an injective holomorphic map ${\Phi }_j:U_j\to \mathbb{D}$ for each domain. Then ${\Phi }_j\circ f$ is a bounded entire function and is constant by Liouville's theorem; injectivity of ${\Phi }_j$ forces $f$ to be constant.

For the upper half-plane, use the Cayley transform between the half-plane and diskIts denominator is nonzero there, and $|z+i{|}^2-|z-i{|}^2=4\mathrm{Im}z>0$, so its image lies in the unit disk. The determinant is nonzero, hence it is injective and conformal.

On the plane slit along the nonnegative real axis, choose the branchThis is the holomorphic square root obtained from the corresponding logarithm branch. Its argument lies in $(0,\pi )$, so it maps $U_2$ bijectively onto the upper half-plane, with inverse $w\mapsto w^2$. Its derivative is $1/(2S(z))\ne 0$. Thereforeis an injective conformal map of $U_2$ onto the unit disk.

For the complement of the closed interval, first useIts only finite pole is the excluded point one, and it is injective with derivative $1/(1-z{)}^2\ne 0$. If $M(z)=t\ge 0$, solving gives $z=t/(1+t)\in [0,1)$, which is excluded. Hence $M(U_3)\subset U_2$, andis the required injective conformal map into the disk. More precisely, $M(U_3)=U_2\setminus \{-1\}$ and ${\Phi }_3(U_3)=\mathbb{D}\setminus \{0\}$, since $S(-1)=i$. No claim that the interval complement is conformally equivalent to the whole disk is required.

Applying Liouville to the three bounded compositions givesThis illustrates Liouville rigidity from a bounded holomorphic embedding: an explicit injective bounded coordinate is enough.