Let \(\Omega \subset \CC\) be the region inside the unit circle \(\abs{z} = 1\) and outside the circle \(\abs{z - \frac{1}{4}} = \frac{1}{4}\). Find a one-to-one conformal map of \(\Omega\) onto an annulus \(r < \abs{z} < 1\) for an appropriate value of \(r\).
Solution
Put \[ \rho=2-\sqrt3. \] Then \[ 0<\rho<1, \qquad \rho^{-1}=2+\sqrt3, \qquad 1+\rho^2=4\rho. \] Define \[ T(z)=\frac{z-\rho}{z-\rho^{-1}}. \]
1. For every \(z\neq\rho^{-1}\),\[ \abs{T(z)}^2-\rho^2 = \frac{(1-\rho^2)(\abs z^2-1)} {\abs{z-\rho^{-1}}^2}. \]
Since \(\rho\rho^{-1}=1\), \[ \begin{aligned} &\abs{z-\rho}^2 - \rho^2\abs{z-\rho^{-1}}^2\\ &\qquad= (1-\rho^2)\abs z^2+\rho^2-1\\ &\qquad= (1-\rho^2)(\abs z^2-1). \end{aligned} \] Dividing by \[ \abs{z-\rho^{-1}}^2 \] gives the identity.
2. For every \(z\neq\rho^{-1}\),\[ \abs{T(z)}^2-\rho^4 = \frac{(1-\rho^4) \left(\abs z^2-\frac12\operatorname{Re}z\right)} {\abs{z-\rho^{-1}}^2}. \]
Expanding gives \[ \begin{aligned} &\abs{z-\rho}^2 - \rho^4\abs{z-\rho^{-1}}^2\\ &\qquad= (1-\rho^4)\abs z^2 - 2\rho(1-\rho^2)\operatorname{Re}z. \end{aligned} \] The relation \[ 1+\rho^2=4\rho \] implies \[ 2\rho(1-\rho^2) = \frac12(1-\rho^4). \] Substitution yields \[ \abs{z-\rho}^2 - \rho^4\abs{z-\rho^{-1}}^2 = (1-\rho^4) \left(\abs z^2-\frac12\operatorname{Re}z\right). \] Dividing by the denominator proves the claim.
3. The map \(T\) sends \(\Omega\) into the annulus\[ \rho^2<\abs w<\rho. \]
If \(z\in\Omega\), then \[ \abs z<1. \] Since \(1-\rho^2>0\), step 1 gives \[ \abs{T(z)}<\rho. \]
The other defining inequality is \[ \abs{z-\frac14}>\frac14. \] Squaring and simplifying gives \[ \abs z^2-\frac12\operatorname{Re}z>0. \] Since \(1-\rho^4>0\), step 2 therefore gives \[ \abs{T(z)}>\rho^2. \]
4. The map\[ T:\Omega\longrightarrow \{w\in\CC:\rho^2<\abs w<\rho\} \]is a conformal bijection.
The map \(T\) is a Möbius transformation. Its pole \[ \rho^{-1}=2+\sqrt3 \] lies outside the unit disk, hence outside \(\Omega\), and \[ T'(z) = \frac{\rho-\rho^{-1}}{(z-\rho^{-1})^2} \neq 0 \] on \(\Omega\).
For surjectivity, let \[ \rho^2<\abs w<\rho. \] Since \(\abs w<1\), one has \(w\neq1\), and the Möbius inverse \[ T^{-1}(w) = \frac{\rho^{-1}w-\rho}{w-1} \] is defined. Apply the identities in steps <1>1–<1>2 to \[ z=T^{-1}(w). \] The inequality \(\abs w<\rho\) forces \(\abs z<1\), while \(\abs w>\rho^2\) forces \[ \abs z^2-\frac12\operatorname{Re}z>0, \] equivalently \[ \abs{z-\frac14}>\frac14. \] Thus \(z\in\Omega\). Hence \(T\) is onto the displayed annulus; injectivity follows from the Möbius inverse.
5. A one-to-one conformal map of \(\Omega\) onto an annulus\[ r<\abs w<1 \]is\[ \boxed{ F(z) = \frac1{\rho} \frac{z-\rho}{z-\rho^{-1}}, \qquad r=\rho=2-\sqrt3. } \]
By step 4, \[ \rho^2<\abs{T(z)}<\rho. \] Dividing by \(\rho\) gives \[ \rho<\abs{F(z)}<1. \] Multiplication by the nonzero constant \(\rho^{-1}\) preserves conformality and injectivity, so \(F\) is the required conformal bijection.
6. Q.E.D.
Step 5 gives both the map and the required inner radius.