For a simply connected pointed region \((U,p)\), write its conformal radius as \(|F'(0)|\), where \[ F:(\Delta,0)\longrightarrow(U,p) \] is a Riemann map.
Find the conformal radii of:
\((\mathbb H,i)\);
\((\{z:-1<\operatorname{Re}z<1\},0)\);
\((\{z\in\mathbb H:-\pi<\operatorname{Re}z<\pi\},i)\);
\((\widehat{\mathbb C}\setminus[-2,2],\infty)\); and
\((S_\alpha,r)\) for \(r>0\), where \[ S_\alpha=\{z:\arg z\in(-\alpha,\alpha)\}. \]
Solution
The conformal radii are \[ \boxed{ \begin{array}{c|c} \text{pointed region}&\text{conformal radius}\\ \hline (\mathbb H,i)&2\\[1mm] (\{-1<\operatorname{Re}z<1\},0)&4/\pi\\[1mm] (\{z\in\mathbb H:-\pi<\operatorname{Re}z<\pi\},i) &4\tanh(1/2)\\[1mm] (\widehat{\mathbb C}\setminus[-2,2],\infty)&1\\[1mm] (S_\alpha,r)&4\alpha r/\pi. \end{array}} \] For the pointed region at infinity, the derivative is measured in the standard local coordinate \(w=1/z\) at \(\infty\).
1. If \(\phi:(U,p)\to(V,q)\) is conformal, then\[ r_U(p)=\frac{r_V(q)}{|\phi'(p)|}, \]with derivatives interpreted in local coordinates.
Let \[ F:(\Delta,0)\to(U,p) \] be a Riemann map. Then \(\phi\circ F\) is a Riemann map from \((\Delta,0)\) to \((V,q)\), so \[ r_V(q)=|(\phi\circ F)'(0)| =|\phi'(p)|\,|F'(0)| =|\phi'(p)|r_U(p). \] Rearranging gives the formula.
2. For \((\mathbb H,i)\),\[ r_{\mathbb H}(i)=2. \]
The Cayley map \[ F(z)=i\frac{1+z}{1-z} \] maps \(\Delta\) conformally onto \(\mathbb H\) and satisfies \(F(0)=i\). Its derivative is \[ F'(z)=\frac{2i}{(1-z)^2}, \] so \[ r_{\mathbb H}(i)=|F'(0)|=2. \] More generally, affine automorphisms of \(\mathbb H\) give \[ r_{\mathbb H}(x+iy)=2y. \]
3. For the vertical strip\[ V=\{z:-1<\operatorname{Re}z<1\}, \]one has\[ r_V(0)=\frac4\pi. \]
The function \[ F(z)=\frac{4i}{\pi}\operatorname{artanh}z =\frac{2i}{\pi}\log\frac{1+z}{1-z} \] maps \(\Delta\) conformally onto \(V\): the function \(\operatorname{artanh}z\) maps the disk onto the horizontal strip \(|\operatorname{Im}w|<\pi/4\), and multiplication by \(4i/\pi\) rotates and rescales it to \(|\operatorname{Re}z|<1\). Also \(F(0)=0\) and \[ F'(0)=\frac{4i}{\pi}. \] Hence \[ r_V(0)=\frac4\pi. \]
4. For\[ W=\{z\in\mathbb H:-\pi<\operatorname{Re}z<\pi\}, \]one has\[ r_W(i)=4\tanh(1/2). \]
The map \[ \phi(z)=\sin(z/2) \] maps \(W\) conformally onto \(\mathbb H\). Indeed, after the affine change \(w=(z+\pi)/2\), the domain becomes the half-strip \[ 0<\operatorname{Re}w<\pi, \qquad \operatorname{Im}w>0, \] which \(w\mapsto-\cos w\) maps conformally onto \(\mathbb H\); the identity \[ -\cos((z+\pi)/2)=\sin(z/2) \] gives the stated map to \(\mathbb H\).
At \(p=i\), \[ \phi(i)=i\sinh(1/2), \qquad |\phi'(i)|=\frac12\cosh(1/2). \] By 2, \[ r_{\mathbb H}(\phi(i))=2\sinh(1/2). \] Therefore 1 gives \[ r_W(i) =\frac{2\sinh(1/2)}{\frac12\cosh(1/2)} =4\tanh(1/2). \]
5. For\[ X=\widehat{\mathbb C}\setminus[-2,2] \]with basepoint \(\infty\),\[ r_X(\infty)=1 \]in the local coordinate \(w=1/z\) at infinity.
The Joukowski map \[ F(\zeta)=\zeta+\frac1\zeta \] is conformal from the punctured disk \(0<|\zeta|<1\) onto \(\widehat{\mathbb C}\setminus[-2,2]\), and it extends by \(F(0)=\infty\). Injectivity follows because \[ F(\zeta_1)=F(\zeta_2) \] implies either \(\zeta_1=\zeta_2\) or \(\zeta_1\zeta_2=1\), and the latter is impossible when both moduli are less than \(1\).
Using the target coordinate \(w=1/z\) near infinity, \[ (w\circ F)(\zeta) =\frac1{\zeta+\zeta^{-1}} =\frac{\zeta}{1+\zeta^2}. \] Thus \[ (w\circ F)'(0)=1, \] so the conformal radius is \(1\) in this standard coordinate.
6. For the sector\[ S_\alpha=\{z:\arg z\in(-\alpha,\alpha)\} \]at the positive real point \(r>0\),\[ r_{S_\alpha}(r)=\frac{4\alpha r}{\pi}. \]
Set \[ \kappa=\frac{\pi}{2\alpha}. \] On the sector choose the branch of \(z^\kappa\) with argument in \((-\pi/2,\pi/2)\). Then \[ \phi(z)=iz^\kappa \] maps \(S_\alpha\) conformally onto \(\mathbb H\). At the positive real point \(r\), \[ \phi(r)=ir^\kappa, \qquad |\phi'(r)|=\kappa r^{\kappa-1}. \] Using 2 and then 1, \[ \begin{aligned} r_{S_\alpha}(r) &=\frac{2r^\kappa}{\kappa r^{\kappa-1}}\\ &=\frac{2r}{\kappa} =\frac{4\alpha r}{\pi}. \end{aligned} \]