Past exam of the complex analysis: conformal maps and geometry course of the University of Oxford

Source: released examination PDF, pages 349–352. Solutions follow the numbering of the 6 June 2023 paper. In 2(d), the depicted domain is interpreted as a simply connected Jordan domain and the separating curves are boundary-to-boundary crosscuts. These topological details matter. The square-root construction in 1(b) is the odd transform of a general normalized univalent map, rather than one fixed extremal Koebe function.

For $a\in \mathbb{D}$, the automorphism of the unit disk $T_a(z)=(z-a)/(1-\bar{a}z)$ sends $a$ to zero. The holomorphic function $F=T_{f(a)}\circ f\circ T_a^{-1}$ maps the unit disc into itself and satisfies $F(0)=0$. The Schwarz lemma therefore gives $|F(w)|\le |w|$. Evaluate at $w=T_a(b)$ to obtain the Schwarz-Pick theorem:The denominators do not vanish because both their products have modulus less than one.

The Koebe growth theorem states that every normalized univalent function $\varphi \in \mathcal{S}$ satisfiesBoth bounds are sharp. Rotations of the Koebe function, ${\varphi }_{\theta }(z)=z/(1-e^{i\theta }z{)}^2$, attain them along suitable radial directions.

The quotient $q(w)=\varphi (w)/w$ extends to a holomorphic function at zero with value one. Since $\varphi$ is injective and vanishes at zero, $q$ has no zero in the unit disc. A holomorphic logarithm $\ell$ exists there, with $\ell (0)=0$. Define the odd square-root transform of a normalized univalent function byIt satisfies $h(z{)}^2=\varphi (z^2)$, $h(0)=0$, $h^{\prime }(0)=1$ and $h(-z)=-h(z)$. To check injectivity, $h(z)=h(w)$ gives $\varphi (z^2)=\varphi (w^2)$, hence $z^2=w^2$. If $w=-z$, oddness gives $h(z)=0$, so $z=w=0$. Thus the two arguments always coincide, and $h\in \mathcal{S}$. Choosing the square root by its derivative at zero removes its sign ambiguity.

For an injective holomorphic function, the real Jacobian determinant is $|h^{\prime }{|}^2$. The area formula for a univalent power series, obtained by change of variables, is thereforeIndeed, insert $h^{\prime }(\rho e^{i\theta })={\sum }_{n\ge 1}n{c}_n{\rho }^{n-1}{e}^{i(n-1)\theta }$. Orthogonality eliminates the cross terms in the angular integral, and radial integration gives $2\pi n^2{\int }_0^r{\rho }^{2n-1}d\rho =\pi n{r}^{2n}$. The power series converges uniformly on each smaller closed disk, justifying the calculation for every $r<1$.

Since $|h(z){|}^2=|\varphi (z^2)|$, the Koebe growth theorem gives $|h(z)|\le |z|/(1-|z{|}^2)$. Hence $h(r\mathbb{D})$ lies in the disk of radius $r/(1-r^2)$, so

Because $|h^2|=|h{|}^2$, the Parseval identity for the Taylor series givesDifferentiate this power series for $r<1$ and use iii:Since $M(0)=0$, integration yields the integral mean bound for an odd square-root transform,

Use $h(\sqrt{r}{e}^{i\theta }{)}^2=\varphi (r{e}^{2i\theta })$. As $\theta$ runs around the circle, the doubled angle goes around it twice, so iv givesThe Cauchy integral formula for the Taylor coefficients now impliesFor $n\ge 2$, choosing $r=(n-1)/n$ gives the linear coefficient bound for normalized univalent functions,The last inequality follows from $\log (1+x)\le x$ for $x>-1$.

By conformal invariance of extremal length, work first on $1<|w|<R$. For a conformal metric $\rho$, let $L_{\rho }$ be the infimum of lengths of curves joining the two circles, and $A_{\rho }$ its area. Each radial segment is admissible, so the Cauchy-Schwarz inequality gives, for every angle,Integrating in angle gives $2\pi L_{\rho }^2\le A_{\rho }\log R$. Conversely, $\rho (w)=1/|w|$ has shortest joining length $\log R$ and area $2\pi \log R$, attaining the bound. Thus the conformal modulus of an annulus givesThe second formula is the pullback of $c/|w|$ by $\psi$. Pullback preserves curve lengths and metric area, so every displayed positive multiple is an extremal metric in $\Omega$.

The symmetry rule for extremal length says that if a domain and its curve family are invariant under reflection, the supremum defining extremal length may be taken over reflection-invariant conformal metrics. More generally, for a conformal or anticonformal involution $T$, use metrics satisfying $T^{\ast }\rho =\rho$.

To see this, replace $\rho$ by ${\rho }_s=(\rho +T^{\ast }\rho )/2$. Every curve in the invariant family and its image have length at least $L_{\rho }$, so $L_{{\rho }_s}\ge L_{\rho }$. Convexity of the square and change of variables give $A_{{\rho }_s}\le A_{\rho }$. Therefore symmetrizing cannot reduce the length-squared to area ratio.

Regard the upper half-disc, with cyclic boundary vertices $-1,0,x,1$, as a conformal quadrilateral. Map it to a rectangle of width $w$ and height $h$, so the boundary arcs $[-1,0]$ and $[x,1]$ become its vertical sides. The rectangle formulas for the conformal modulus of a quadrilateral giveThe remaining two boundary arcs, $(0,x)$ and the semicircle, become the horizontal sides. Thus these are precisely the dual joining families in the same quadrilateral.

Use the simply connected domain shown in the diagram. A Riemann mapping theorem map followed by an automorphism of the unit disk sends its marked points to $0$ and $x\in (0,1)$. Split this disc along its real diameter. Map the upper conformal quadrilateral from c to a rectangle $[0,w]\times [0,h]$, with $[-1,0]$ and $[x,1]$ as vertical sides and $(0,x)$ as its bottom. The lower half is a second identical rectangle. The full disc is obtained by gluing the two rectangles along their two vertical sides and their bottom sides; the two top sides form the original outer boundary.

In this doubled rectangle extremal-length duality, loops enclosing both marked bottom corners have shortest flat length $2w$. They must meet both vertical seams, and a closed curve's total horizontal travel is at least $2w$; the doubled horizontal segments attain it. Crosscuts separating the two corners must meet the bottom seam, and run from outer boundary back to outer boundary. Their total vertical travel is at least $2h$, attained by doubled vertical segments. The total flat area is $2wh$.

These flat metrics are extremal, not merely lower bounds. For an arbitrary metric, average the lengths of the doubled horizontal segments over their height and apply the Cauchy-Schwarz inequality to get $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Averaging the doubled vertical segments gives $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. ConsequentlyThe conformal metric on the doubled rectangles pulls back to an admissible metric on the original punctured domain; the two marked corner points have zero area. Here ${\Gamma }^{\ast }$ consists of crosscuts with endpoints on the outer boundary, as drawn. Boundedness alone, without the illustrated domain topology and the specified curve families, would not justify this construction.

The radial harmonic function with value zero at $|z|=r$ and one at $|z|=1$ isIndeed $\log |z|$ is harmonic away from zero, and the displayed expression has the required boundary values. Uniqueness of the Dirichlet problem identifies it as the harmonic measure of an annulus.

The concentric normalization of an eccentric annulus here usesThis Möbius transformation preserves the unit circle and sends the inner circle's real endpoints $0,4/5$ to $-1/2,1/2$. Its image circle is symmetric about the real axis and hence is $|w|=1/2$. Equivalently, expanding $|T(z){|}^2=1/4$ gives $|z{|}^2=(4/5)\mathrm{Re}z$, precisely the inner-circle equation. Thus $T$ sends the domain to $1/2<|w|<1$. Conformal invariance of harmonic measure and i giveThe zero of $T$ lies in the removed inner disk, so this logarithm is harmonic throughout $\Omega$.

For $0<|z|<R$, set $w=R^2/\bar{z}$, which lies outside the closed disk. On $|\zeta |=R$, direct algebra gives the Poisson kernel identityUsing $d\theta =d\zeta /(i\zeta )$, the proposed Poisson formula for a holomorphic function becomesThe first term is $f(z)$ by the Cauchy integral formula; the second vanishes by the Cauchy integral theorem, since its pole lies outside the disk. At $z=0$ the kernel is one, and the same identity is the mean value property for holomorphic functions. Therefore

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

Fix $r<1$ and construct $g_n$ from the boundary values of $u_n$ on the circle $|z|=r$ as in ii. By the assumed locally uniform convergence, those boundary values converge uniformly to $u$. For each $s<r$, the kernel is bounded on $|z|\le s$, so the same integral shows that $g_n$ converges uniformly there toThe Weierstrass convergence theorem makes $g$ holomorphic. Since $\mathrm{Re}{g}_n=u_n$, passing to the limit gives $u=\mathrm{Re}g$ on $|z|<r$. The real part of a holomorphic function is harmonic. As $r<1$ was arbitrary, $u$ is harmonic throughout the unit disc. This proves the locally uniform limit of harmonic functions theorem without presuming convergence of the derivatives of $u_n$.