Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 24F (Riemann Surfaces)

The uniformization theorem says that every simply connected Riemann surface is conformally equivalent to exactly one of the Riemann sphere, the complex plane, and the unit disc. A proper simply connected plane domain is noncompact, so it is not the sphere. It cannot be uniformized by the plane: after choosing a point outside the domain, a branch of a square root and then a Möbius transformation would produce a nonconstant bounded entire function, contradicting the Liouville theorem. It is therefore conformally equivalent to the disc, which is the Riemann mapping theorem.

Use the principal branch $\mathrm{Log}z=\log |z|+i\arg z$, with $-\pi <\arg z<\pi$. Thenis a conformal equivalence from the slit plane to the vertical strip. Its imaginary part tends to $-\infty$ as $z\to 0$ and to $+\infty$ as $z\to \infty$.

The map $h(z)=e^{-iz}$ is a conformal equivalence from $E$ to the plane slit along the nonpositive real axis. Conjugating $\varphi$ by $h$ gives an automorphism of the slit plane whose two prime ends at $0$ and $\infty$ are fixed. Taking the principal square root identifies the slit plane with the right half-plane; the automorphisms of a half-plane that fix $0$ and $\infty$ are precisely $w\mapsto aw$ with $a>0$. Undoing the square root changes only the positive scale, sofor some $a>0$. Since $E$ is connected and both sides use the same branch,Thus $\varphi$ is translation by a purely imaginary number.

If $f_0(z)=i\mathrm{Log}z$ is the projection from part (b), then any other Mercator projection $f$ gives an automorphism $f\circ f_0^{-1}$ of $E$ preserving its two ends. Part (c) therefore yieldsEvery map in this family plainly has the required limiting behaviour.