Let \(f:\Delta\to\mathbb C\) be univalent with \[ f(0)=0, \qquad f'(0)=1, \] and suppose \[ f(iz)=if(z) \] for every \(z\in\Delta\). Show that \[ B\left(0,\frac1{\sqrt2}\right)\subseteq f(\Delta). \]
Solution
It is enough to prove that every omitted point \(p\notin f(\Delta)\) satisfies \[ |p|\ge\frac1{\sqrt2}. \]
1. The exterior transform\[ F(\zeta)=\frac1{f(1/\zeta)} \]belongs to \(\Sigma\) and satisfies\[ F(i\zeta)=iF(\zeta). \]
The symmetry \(f(iz)=if(z)\) implies that the Taylor expansion of \(f\) has only powers congruent to \(1\) modulo \(4\): \[ f(z)=z+a_5z^5+a_9z^9+\cdots. \] Hence \[ F(\zeta) =\zeta+\frac{b_3}{\zeta^3}+\frac{b_7}{\zeta^7}+\cdots, \] so \(F\) has the required normalization at infinity. It is univalent because \(f\) is univalent. Also \[ F(i\zeta) =\frac1{f(1/(i\zeta))} =\frac1{f(-i/\zeta)} =\frac1{-i f(1/\zeta)} =iF(\zeta). \] In particular \(F(-\zeta)=-F(\zeta)\).
2. If \(p\notin f(\Delta)\) and\[ w=\frac1p, \]then both \(w\) and \(-w\) are omitted by \(F\).
Since \(0=f(0)\), an omitted point \(p\) is nonzero. If \(F(\zeta)=w=1/p\), then \[ f(1/\zeta)=p, \] contrary to the choice of \(p\). Thus \(w\) is omitted. Since \(F\) is odd, \(-w\) is omitted as well.
3. The function\[ H(\zeta)=\sqrt{F(\zeta)^2-w^2}, \qquad H(\zeta)\sim\zeta\quad(\zeta\to\infty), \]is a normalized exterior univalent map, and its first area-theorem coefficient is\[ c_1=-\frac{w^2}{2}. \]
By 2, the function \(F^2-w^2\) has no zero on the exterior disk. More precisely, \[ q(\zeta)=\frac{F(\zeta)^2-w^2}{\zeta^2} \] is zero-free and, in the coordinate \(u=1/\zeta\), extends holomorphically across \(u=0\) with value \(q(\infty)=1\). Thus \(q\) is a zero-free holomorphic function on the simply connected disk \(|u|<1\), so it has a holomorphic square root \(Q\) normalized by \(Q(\infty)=1\). Setting \[ H(\zeta)=\zeta Q(\zeta) \] gives the asserted branch and \(H(\zeta)/\zeta\to1\).
To check univalence, suppose \(H(\zeta_1)=H(\zeta_2)\). Squaring gives \[ F(\zeta_1)^2=F(\zeta_2)^2, \] so either \(F(\zeta_1)=F(\zeta_2)\) or \(F(\zeta_1)=-F(\zeta_2)\). In the first case univalence of \(F\) gives \(\zeta_1=\zeta_2\). In the second, oddness of \(F\) gives \(\zeta_1=-\zeta_2\).
The normalized square root is also odd: both \(H(-\zeta)\) and \(-H(\zeta)\) square to the same function and are asymptotic to \(-\zeta\) at infinity. Hence if \(\zeta_1=-\zeta_2\) and \(H(\zeta_1)=H(\zeta_2)\), then \(H(\zeta_1)=0\), impossible because \(F^2-w^2\) is zero-free. Thus \(H\) is univalent.
Finally, since \[ F(\zeta)=\zeta+O(\zeta^{-3}), \] we have \[ \begin{aligned} H(\zeta) &=\sqrt{\zeta^2-w^2+O(\zeta^{-2})}\\ &=\zeta-\frac{w^2}{2\zeta}+O(\zeta^{-3}). \end{aligned} \] Thus \(H\in\Sigma\) and its coefficient of \(1/\zeta\) is \(c_1=-w^2/2\).
4. The area theorem gives the required disk.
For every map in \(\Sigma\), the area theorem implies \[ |c_1|\le1. \] Applying this to \(H\) gives \[ \frac{|w|^2}{2}\le1, \qquad\text{so}\qquad |w|\le\sqrt2. \] Since \(w=1/p\), \[ |p|\ge\frac1{\sqrt2}. \] Thus no point of modulus less than \(1/\sqrt2\) can be omitted by \(f\), and therefore \[ B\left(0,\frac1{\sqrt2}\right)\subseteq f(\Delta). \]