Let \[ f(z)=z+\sum_{n=1}^{\infty}\frac{b_n}{z^n} \] be a univalent map from \(\widehat{\mathbb C}\setminus\Delta\) to \(\widehat{\mathbb C}\) fixing \(\infty\) with derivative \(1\) there. If \(w\) is not in the image of \(f\), prove that \[ |w|\le2. \]
Hint: apply the area-theorem bound \(|b_1|\le1\) to a normalized square root of \(f(z^2)-w\).
Solution
Let \[ G(z)=\sqrt{f(z^2)-w} \] with the branch normalized by \(G(z)/z\to1\) at infinity. Then \(G\in\Sigma\) and its first negative Laurent coefficient is \(-w/2\). The area theorem gives the result.
1. The function \(f(z^2)-w\) has a holomorphic square root on\(\widehat{\mathbb C}\setminus\overline\Delta\) normalized by\[ G(z)=z+O(z^{-1}). \]
Because \(w\) is omitted by \(f\), the function \[ Q(z)=f(z^2)-w \] has no zeros for \(|z|>1\). Near infinity, \[ Q(z)=z^2-w+O(z^{-2}) =z^2\left(1-wz^{-2}+O(z^{-4})\right). \] Thus \[ \frac{Q(z)}{z^2}\longrightarrow1 \qquad(z\to\infty). \] The nonvanishing function \(Q(z)/z^2\) has winding number zero around \(0\) on a large circle, and hence on every circle in the exterior domain by homotopy through nonvanishing values. It therefore has a holomorphic logarithm there. Define \[ G(z) =z\exp\left(\frac12\log\frac{Q(z)}{z^2}\right), \] choosing the logarithm tending to \(0\) at infinity. Then \(G^2=Q\) and \(G(z)/z\to1\).
2. The map \(G\) is univalent on the exterior disk.
Suppose \(G(z_1)=G(z_2)\). Squaring gives \[ f(z_1^2)-w=f(z_2^2)-w. \] Since \(f\) is univalent, \[ z_1^2=z_2^2, \] so \(z_2=\pm z_1\).
Because \(Q(-z)=Q(z)\), the quotient \[ \frac{G(-z)}{G(z)} \] has square \(1\) and is holomorphic on the connected exterior domain. Hence it is constant. From the normalization at infinity, \[ \frac{G(-z)}{G(z)}\longrightarrow-1, \] so \[ G(-z)=-G(z). \] Also \(G\) never vanishes because \(Q\) never vanishes. Therefore \(G(z_1)=G(-z_1)\) is impossible, and we must have \(z_2=z_1\). Thus \(G\) is univalent.
3. The Laurent expansion of \(G\) begins\[ G(z)=z-\frac{w}{2z}+O(z^{-3}). \]
Using \[ f(z^2) =z^2+\frac{b_1}{z^2}+\frac{b_2}{z^4}+\cdots, \] we have \[ Q(z) =z^2\left(1-wz^{-2}+b_1z^{-4}+O(z^{-6})\right). \] Hence \[ \begin{aligned} G(z) &=z\left(1-wz^{-2}+O(z^{-4})\right)^{1/2}\\ &=z\left(1-\frac w2z^{-2}+O(z^{-4})\right)\\ &=z-\frac{w}{2z}+O(z^{-3}). \end{aligned} \] Thus, if \[ G(z)=z+\sum_{n\ge1}\frac{c_n}{z^n}, \] then \[ c_1=-\frac w2. \]
4. The area theorem implies \(|w|\le2\).
By 1 and 2, \(G\in\Sigma\). The area theorem gives \[ |c_1|\le1. \] Using 3, \[ \frac{|w|}{2}=|c_1|\le1. \] Therefore \[ \boxed{|w|\le2}. \]