Find a conformal map
from \(\{ z: |z - 1/2| > 1/2, \text{Re}(z)>0 \}\) to \(\mathbb H\)
from \(\{ z: |z - 1/2| > 1/2, |z| <1 \}\) to \(\mathbb D\)
from the intersection of the disk \(|z + i| < \sqrt{2}\) with \({\mathbb H}\) to \({\mathbb D}\).
from \({\mathbb D} \backslash [a, 1)\) to \({\mathbb D} \backslash [0, 1)\) (\(0<a<1)\).
Short solution possible using Blaschke factor.
from \(\{ z: |z| < 1, \text{Re}(z) > 0 \} \backslash (0, 1/2]\) to \(\mathbb H\).
Solution
Goal: Construct explicit conformal mappings (biholomorphisms) for each of the five given domains.
Part 1: Map from \(\Omega_1 = \{ z \in \mathbb{C} : |z - 1/2| > 1/2, \text{Re}(z)>0 \}\) to \(\mathbb{H} = \{w : \text{Im}(w) > 0\}\)
1. The inversion \(w_1 = \frac{1}{z}\) maps \(\Omega_1\) conformally onto the vertical strip \(S_1 = \{w_1 \in \mathbb{C} : 0 < \text{Re}(w_1) < 1\}\).
1.1. For \(z = x+iy \neq 0\), \(\text{Re}(1/z) = \frac{x}{x^2+y^2}\).Proof: \(\frac{1}{x+iy} = \frac{x-iy}{x^2+y^2}\).
1.2. \(\text{Re}(z) > 0 \iff x > 0 \iff \text{Re}(1/z) > 0\).Proof: \(x^2+y^2 > 0\) for all \(z \neq 0\), so \(x > 0 \iff \frac{x}{x^2+y^2} > 0\).
1.3. \(|z - 1/2| > 1/2 \iff (x - 1/2)^2 + y^2 > 1/4 \iff x^2 - x + y^2 > 0 \iff x^2+y^2 > x \iff \frac{x}{x^2+y^2} < 1 \iff \text{Re}(1/z) < 1\).Proof: Direct algebraic manipulation of the disc boundary equation.
1.4. The Möbius transformation \(z \mapsto 1/z\) is biholomorphic from \(\Omega_1\) to \(S_1\).Proof: Möbius transformations are conformal automorphisms of the Riemann sphere, and this map is a bijection between the open sets \(\Omega_1\) and \(S_1\).
1.5. Q.E.D.
2. The transformation \(w_2 = i\pi w_1\) maps \(S_1\) onto the horizontal strip \(\{w_2 \in \mathbb{C} : 0 < \text{Im}(w_2) < \pi\}\).
2.1. For \(w_1 = u+iv\), \(w_2 = i\pi(u+iv) = -\pi v + i\pi u\).Proof: Multiplication by \(i\pi\).
2.2. \(0 < u < 1 \iff 0 < \text{Im}(w_2) < \pi\), while \(\text{Re}(w_2) = -\pi v \in \mathbb{R}\).Proof: Follows from definition of \(w_2\).
2.3. Q.E.D.
3. The exponential map \(w \mapsto e^{w_2} = \exp\left(\frac{i\pi}{z}\right)\) maps the horizontal strip conformally onto the upper half-plane \(\mathbb{H}\).
3.1. The exponential function maps the strip \(\{ \xi + i\eta : \xi \in \mathbb{R}, 0 < \eta < \pi \}\) biholomorphically onto \(\mathbb{H} = \{ r e^{i\theta} : r > 0, 0 < \theta < \pi \}\).Proof: Standard property of the complex exponential function.
3.2. Thus, the composite map \(f_1(z) = \exp\left(\frac{i\pi}{z}\right)\) is a conformal map from \(\Omega_1\) to \(\mathbb{H}\).Proof: Composition of biholomorphisms is a biholomorphism.
3.3. Q.E.D.
Part 2: Map from \(\Omega_2 = \{ z : |z - 1/2| > 1/2, |z| < 1 \}\) to \(\mathbb{D}\)
4. Apply the inversion \(w_1 = \frac{1}{z-1}\).
4.1. The boundary circle \(|z| = 1\) passes through \(z=1\), so under \(w_1 = \frac{1}{z-1}\), it maps to a straight line.Since \(\text{Re}\left(\frac{1}{e^{i\theta}-1}\right) = \text{Re}\left(\frac{e^{-i\theta}-1}{|e^{i\theta}-1|^2}\right) = \frac{\cos\theta - 1}{2-2\cos\theta} = -\frac{1}{2}\), the image of \(|z| < 1\) is the half-plane \(\text{Re}(w_1) < -\frac{1}{2}\).Proof: Standard property of inversion mapping circles through the pole to straight lines.
4.2. The inner circle \(|z - 1/2| = 1/2\) also passes through \(z=1\).For points on this circle, \(z = \frac{1}{2} + \frac{1}{2}e^{i\theta}\), so \(z-1 = -\frac{1}{2}(1 - e^{i\theta})\).Then \(\text{Re}\left(\frac{1}{z-1}\right) = \text{Re}\left(\frac{-2}{1-e^{i\theta}}\right) = -2 \cdot \frac{1}{2} = -1\).Proof: Calculation of real part of \(\frac{-2}{1 - \cos\theta - i\sin\theta}\).
4.3. Since \(|z-1/2| > 1/2\), the region \(\Omega_2\) maps conformally onto the vertical strip \(S_2 = \{w_1 : -1 < \text{Re}(w_1) < -1/2\}\).Proof: By continuity and connectedness, evaluating at the test point \(z = -1/2 \in \Omega_2\) gives \(w_1 = \frac{1}{-3/2} = -2/3 \in (-1, -1/2)\).
4.4. Q.E.D.
5. Map the strip \(S_2\) to \(\mathbb{D}\).
5.1. Shift and scale \(w_1\) to the standard strip: \(w_2 = 2\pi\left(w_1 + 1\right) = 2\pi\left(\frac{1}{z-1} + 1\right) = \frac{2\pi z}{z-1}\).When \(w_1 \in (-1, -1/2)\), \(\text{Re}(w_2) \in (0, \pi)\).Proof: Linear change of variable maps \((-1, -1/2)\) to \((0, \pi)\).
5.2. Map to horizontal strip: \(w_3 = i w_2 = \frac{2\pi i z}{z-1}\), which has \(0 < \text{Im}(w_3) < \pi\).Proof: \(\text{Im}(i w_2) = \text{Re}(w_2) \in (0, \pi)\).
5.3. Exponentiate: \(w_4 = e^{w_3} = \exp\left(\frac{2\pi i z}{z-1}\right)\) maps onto the upper half-plane \(\mathbb{H}\).Proof: Standard exponential mapping of a horizontal strip of width \(\pi\).
5.4. Cayley transform to \(\mathbb{D}\): \(f_2(z) = \frac{w_4 - i}{w_4 + i} = \frac{\exp\left(\frac{2\pi i z}{z-1}\right) - i}{\exp\left(\frac{2\pi i z}{z-1}\right) + i}\) is a conformal map from \(\Omega_2\) to \(\mathbb{D}\).Proof: The Cayley transform \(w \mapsto \frac{w-i}{w+i}\) maps \(\mathbb{H}\) biholomorphically onto \(\mathbb{D}\).
5.5. Q.E.D.
Part 3: Map from \(\Omega_3 = \{z \in \mathbb{H} : |z + i| < \sqrt{2}\}\) to \(\mathbb{D}\)
6. Identify the geometry of \(\Omega_3\).
6.1. The region is bounded by the real line \(\text{Im}(z) = 0\) and the circular arc \(|z+i| = \sqrt{2}\) in the upper half-plane.Proof: By definition of \(\Omega_3\).
6.2. The circle \(|z+i|^2 = x^2 + (y+1)^2 = 2\) intersects \(y=0\) at \(x^2 + 1 = 2 \implies x = \pm 1\).The vertices of the circular lune are \(z_1 = -1\) and \(z_2 = 1\).Proof: Solving \(x^2+(0+1)^2=2\).
6.3. The circle \(|z+i|=\sqrt{2}\) meets the real axis at an angle: the normal to the circle at \(z=1\) is \(1 - (-i) = 1+i\) (angle \(\pi/4\)), while the real axis has normal \(i\) (angle \(\pi/2\)). The internal angle of \(\Omega_3\) at both vertices \(\pm 1\) is \(\pi/4\).Proof: At \(z=1\), the tangent vector to the circle is \(e^{i 3\pi/4}\), which makes an angle of \(\pi/4\) with the real segment \([-1, 1]\).
6.4. Q.E.D.
7. Map the lune \(\Omega_3\) conformally to \(\mathbb{D}\).
7.1. The Möbius transformation \(w_1 = \frac{z+1}{z-1}\) maps \(z = -1 \mapsto 0\) and \(z = 1 \mapsto \infty\).Proof: Definition of Möbius transformation send vertices to \(0\) and \(\infty\).
7.2. Under \(w_1\), the segment \((-1, 1) \subset \mathbb{R}\) maps to the negative real axis \((-\infty, 0)\) because for \(x \in (-1,1)\), \(\frac{x+1}{x-1} < 0\).The circular arc (passing through \(z = (\sqrt{2}-1)i\)) maps to the ray of argument \(-3\pi/4 \equiv 5\pi/4\).Proof: For \(z = (\sqrt{2}-1)i\), \(w_1 = \frac{1+(\sqrt{2}-1)i}{-1+(\sqrt{2}-1)i} = \frac{(1+(\sqrt{2}-1)i)^2}{1+(\sqrt{2}-1)^2} = \frac{2\sqrt{2}-2 + 2(\sqrt{2}-1)i}{4-2\sqrt{2}} = -1 + i\), which has \(\arg(w_1) = 3\pi/4\).The sector formed between the real segment (ray \(\arg = \pi\)) and the arc has opening angle \(\pi/4\), which is the sector \(\{w_1 : 3\pi/4 < \arg(w_1) < \pi\}\).
7.3. Rotate by \(e^{-3\pi i/4}\): \(w_2 = e^{-3\pi i/4} w_1 = e^{-3\pi i/4} \left(\frac{z+1}{z-1}\right)\) maps \(\Omega_3\) to the sector \(S = \{w_2 : 0 < \arg(w_2) < \pi/4\}\).Proof: Rotation by \(-3\pi/4\).
7.4. Power map \(w_3 = w_2^4 = \left(e^{-3\pi i/4} \frac{z+1}{z-1}\right)^4 = - \left(\frac{z+1}{z-1}\right)^4\) maps the sector of angle \(\pi/4\) onto \(\mathbb{H}\).Proof: \(z \mapsto z^4\) multiplies arguments by 4, sending \((0, \pi/4)\) to \((0, \pi) = \mathbb{H}\), and \((e^{-3\pi i/4})^4 = e^{-3\pi i} = -1\).
7.5. Cayley transform: \(f_3(z) = \frac{w_3 - i}{w_3 + i} = \frac{- \left(\frac{z+1}{z-1}\right)^4 - i}{- \left(\frac{z+1}{z-1}\right)^4 + i} = \frac{\left(\frac{z+1}{z-1}\right)^4 + i}{\left(\frac{z+1}{z-1}\right)^4 - i}\) maps \(\Omega_3\) conformally to \(\mathbb{D}\).Proof: Composition of biholomorphisms.
7.6. Q.E.D.
Part 4: Map from \(\mathbb{D} \setminus [a, 1)\) to \(\mathbb{D} \setminus [0, 1)\) (\(0 < a < 1\))
8. Use an automorphism of the unit disc (Blaschke factor) mapping \(a\) to \(0\).
8.1. Define \(\phi_a(z) = \frac{z - a}{1 - az}\).Proof: For \(a \in (-1, 1)\), \(\phi_a\) is an automorphism of \(\mathbb{D}\) (i.e. \(\phi_a \in \text{Aut}(\mathbb{D})\)).
8.2. \(\phi_a(a) = \frac{a-a}{1-a^2} = 0\), and \(\phi_a(1) = \frac{1-a}{1-a} = 1\).Proof: Direct evaluation.
8.3. For \(x \in [a, 1)\), \(\phi_a(x) = \frac{x-a}{1-ax}\) is strictly increasing since \(\phi_a'(x) = \frac{1-a^2}{(1-ax)^2} > 0\), so \(\phi_a([a, 1)) = [0, 1)\).Proof: \(\phi_a\) is a strictly increasing diffeomorphism on \([a, 1)\) with \(\phi_a(a)=0\) and \(\lim_{x\to 1^-} \phi_a(x)=1\).
8.4. Since \(\phi_a\) is a bijection of \(\mathbb{D}\) to \(\mathbb{D}\) and maps the slit \([a,1)\) bijectively onto the slit \([0,1)\), it restricts to a conformal map from \(\mathbb{D} \setminus [a,1)\) onto \(\mathbb{D} \setminus [0,1)\).Proof: Restriction of a biholomorphism to an open subset.
8.5. Q.E.D.
Part 5: Map from \(\Omega_5 = \{ z : |z| < 1, \text{Re}(z) > 0 \} \setminus (0, 1/2]\) to \(\mathbb{H}\)
9. Square map \(w_1 = z^2\).
9.1. The right half-disc \(\{|z|<1, \text{Re}(z)>0\} = \{r e^{i\theta} : 0 < r < 1, -\pi/2 < \theta < \pi/2\}\).Proof: Polar coordinate description of the right half-disc.
9.2. The map \(w_1 = z^2\) sends \(r e^{i\theta} \mapsto r^2 e^{2i\theta}\), which maps the right half-disc conformally to the slit unit disc \(\mathbb{D} \setminus (-1, 0]\).Proof: Angles \(-\pi/2 < \theta < \pi/2\) map bijectively to \(-\pi < 2\theta < \pi\), so the negative real axis is the boundary slit coming from the imaginary axis.
9.3. The slit \((0, 1/2]\) on the positive real axis maps under \(z \mapsto z^2\) to the slit \((0, 1/4]\) on the positive real axis.Proof: For \(x \in (0, 1/2]\), \(x^2 \in (0, 1/4]\).
9.4. Thus, \(w_1 = z^2\) maps \(\Omega_5\) conformally onto \(\mathbb{D} \setminus ( (-1, 0] \cup (0, 1/4] ) = \mathbb{D} \setminus (-1, 1/4]\).Proof: Union of the two slits.
9.5. Q.E.D.
10. Use an automorphism of \(\mathbb{D}\) to center/shift the slit.
10.1. Consider the Möbius transformation \(\psi(w_1) = \frac{w_1 - 1/4}{1 - w_1/4}\).Proof: This is a Blaschke factor mapping \(\mathbb{D} \to \mathbb{D}\).
10.2. \(\psi(1/4) = 0\), \(\psi(-1) = \frac{-1 - 1/4}{1 + 1/4} = \frac{-5/4}{5/4} = -1\), and \(\psi(1) = 1\).Proof: Direct evaluation.
10.3. Since \(\psi\) is monotone increasing on \((-1, 1)\), it maps the slit \((-1, 1/4]\) onto the slit \((-1, 0]\).Proof: \(\psi((-1, 1/4]) = (\psi(-1), \psi(1/4)] = (-1, 0]\).
10.4. Hence \(w_2 = \psi(z^2) = \frac{z^2 - 1/4}{1 - z^2/4} = \frac{4z^2 - 1}{4 - z^2}\) maps \(\Omega_5\) conformally onto the slit disc \(\mathbb{D} \setminus (-1, 0]\).Proof: Composition of \(z \mapsto z^2\) and \(\psi\).
10.5. Q.E.D.
11. Map \(\mathbb{D} \setminus (-1, 0]\) onto the upper half-plane \(\mathbb{H}\).
11.1. The principal square root \(w_3 = \sqrt{w_2}\) maps \(\mathbb{D} \setminus (-1, 0]\) conformally onto the right half-disc \(\{w_3 \in \mathbb{D} : \text{Re}(w_3) > 0\}\).Proof: The principal square root maps the slit plane \(\mathbb{C} \setminus (-\infty, 0]\) to the right half-plane \(\text{Re}(w_3) > 0\), and preserves the condition \(|w_2| < 1 \iff |w_3| < 1\).
11.2. Joukowsky-type transform: \(w_4 = -i \frac{w_3 - 1}{w_3 + 1}\) maps the right half-disc to the first quadrant, or directly use the Joukowsky map \(J(\zeta) = -\frac{1}{2}(\zeta + 1/\zeta)\) which maps the upper half-disc to \(\mathbb{H}\).Rotating the right half-disc to the upper half-disc via \(w_3 \mapsto i w_3\) and applying \(J\) gives: \(f_5(z) = -\frac{1}{2}\left(i\sqrt{\frac{4z^2-1}{4-z^2}} + \frac{1}{i\sqrt{\frac{4z^2-1}{4-z^2}}}\right) = -\frac{i}{2}\left(\sqrt{\frac{4z^2-1}{4-z^2}} - \sqrt{\frac{4-z^2}{4z^2-1}}\right)\).Proof: Composition of biholomorphisms ending in \(\mathbb{H}\).
11.3. Q.E.D.