Past exam of the complex analysis: conformal maps and geometry course of the University of Oxford / 2023 / 1

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 $x>-1$.