Let \(U\subset\mathbb C\) be simply connected, with \[ 0\in U, \qquad 1\notin U. \]
Suppose the conformal radius of \((U,0)\) is \(2\). Is \(U\) uniquely determined?
What if the conformal radius is \(4\)?
Solution
No. For example, both \[ U_1=\{w:\operatorname{Re}w<1\} \qquad\text{and}\qquad U_2=\mathbb C\setminus[1/2,\infty) \] contain \(0\), omit \(1\), and have conformal radius \(2\) at \(0\).
Yes. If the conformal radius is \(4\), then necessarily \[ \boxed{U=\mathbb C\setminus[1,\infty)}. \]
1. Let \(R\) be the conformal radius of \((U,0)\). Then \(R\le4\); if \(R=4\), the normalized Riemann map is forced to be the Koebe extremal.
Choose the Riemann map \[ F:(\Delta,0)\longrightarrow(U,0) \] with \(F'(0)=R>0\), and put \[ f(z)=\frac{F(z)}R=z+a_2z^2+\cdots. \] Then \(f\in\mathcal S\). Since \(1\notin U\), the point \[ p=\frac1R \] is omitted by \(f\).
The coefficient argument that proves the Koebe one-quarter theorem applies to any omitted point \(p\): the map \[ A_p(w)=\frac{w}{1-w/p} \] satisfies \(A_p(0)=0\), \(A_p'(0)=1\), and \(A_p\circ f\in\mathcal S\). Its second coefficient is \[ a_2+\frac1p=a_2+R. \] The elementary coefficient bound \(|a_2|\le2\) therefore gives \[ |a_2|\le2, \qquad |a_2+R|\le2. \] Consequently \[ R=|(a_2+R)-a_2| \le |a_2+R|+|a_2| \le4. \]
If \(R=4\), the closed disks \[ \{|a_2|\le2\}, \qquad \{|a_2+4|\le2\} \] are tangent, so necessarily \[ a_2=-2. \] We now identify the equality case. Define the odd univalent function \[ h(z)=\sqrt{f(z^2)} =z+\frac{a_2}{2}z^3+\cdots =z-z^3+\cdots, \] where the branch is chosen with \(h'(0)=1\). Its exterior transform \[ H(\zeta)=\frac1{h(1/\zeta)} =\zeta+\frac1\zeta+\frac{b_2}{\zeta^2}+\cdots \] lies in \(\Sigma\). By the area theorem, \[ \sum_{n\ge1}n|b_n|^2\le1. \] Here \(b_1=-a_2/2=1\), so equality is already exhausted by the first term and hence \[ b_n=0\qquad(n\ge2). \] Thus \[ H(\zeta)=\zeta+\frac1\zeta. \] Replacing \(\zeta\) by \(1/z\) gives \[ h(z)=\frac{z}{1+z^2}, \] and squaring yields \[ f(t)=\frac{t}{(1+t)^2} \qquad(t\in\Delta). \] Therefore the Riemann map itself is forced to be \[ F(t)=\frac{4t}{(1+t)^2}. \]
The square root defining \(h\) exists: since \(f\) is univalent and \(f(0)=0\), the function \(f(w)/w\) is holomorphic and zero-free on \(\Delta\), hence has a holomorphic square root. Thus \[ h(z)=z\sqrt{\frac{f(z^2)}{z^2}}. \] It is univalent as well. Indeed, \(h(z_1)=h(z_2)\) implies \(f(z_1^2)=f(z_2^2)\), hence \(z_1^2=z_2^2\). If \(z_2=-z_1\), oddness gives \(h(z_1)=-h(z_1)\), hence \(h(z_1)=0\), and then \(z_1=0\) because \(f\) has no zero except at \(0\). Thus \(z_1=z_2\).
2. The map\[ k(z)=\frac{z}{(1+z)^2} \]is a conformal bijection\[ \Delta\longrightarrow\mathbb C\setminus[1/4,\infty). \]
If \(k(z_1)=k(z_2)\), then \[ z_1(1+z_2)^2=z_2(1+z_1)^2, \] so \[ (z_1-z_2)(1-z_1z_2)=0. \] For \(z_1,z_2\in\Delta\) one has \(|z_1z_2|<1\), hence \(z_1=z_2\); thus \(k\) is univalent.
For \(z=e^{i\theta}\ne-1\), \[ k(e^{i\theta}) =\frac{1}{4\cos^2(\theta/2)} \in[1/4,\infty). \] It remains to identify the image. Solving \(k(z)=w\) gives \[ wz^2+(2w-1)z+w=0. \] For \(w\ne0\), the two roots have product \(1\). A root lies on \(S^1\) exactly when \(w=k(e^{i\theta})\in[1/4,\infty)\). Hence, if \[ w\notin[1/4,\infty), \] neither root lies on \(S^1\), and because their moduli have product \(1\), exactly one root lies in \(\Delta\). Conversely, for \(w\in[1/4,\infty)\) both roots lie on \(S^1\), so there is no root in \(\Delta\). The case \(w=0\) has the unique solution \(z=0\). Therefore \[ k(\Delta)=\mathbb C\setminus[1/4,\infty). \]
3. The two radius-\(2\) domains in part (a) are distinct and satisfy all hypotheses.
For \[ U_1=\{w:\operatorname{Re}w<1\}, \] the map \[ F_1(z)=\frac{2z}{1+z} \] is conformal from \(\Delta\) onto \(U_1\): its inverse is \(z=w/(2-w)\), and \[ \left|\frac{w}{2-w}\right|<1 \iff \operatorname{Re}w<1. \] Since \(F_1(0)=0\) and \(F_1'(0)=2\), the conformal radius is \(2\).
By 2, \[ F_2(z)=2k(z)=\frac{2z}{(1+z)^2} \] maps \(\Delta\) conformally onto \[ U_2=\mathbb C\setminus[1/2,\infty). \] Again \(F_2(0)=0\) and \(F_2'(0)=2\). Both domains omit \(1\), but \(U_1\ne U_2\), so radius \(2\) does not determine \(U\).
4. Radius \(4\) determines the domain uniquely.
By 1, the normalized Riemann map must be \[ F(z)=4k(z)=\frac{4z}{(1+z)^2}. \] By 2 its image is \[ 4\bigl(\mathbb C\setminus[1/4,\infty)\bigr) =\mathbb C\setminus[1,\infty). \] Hence this is the only possible \(U\) when the conformal radius equals \(4\).