Let \(\displaystyle{\psi_{\alpha}(z)=\frac{\alpha-z}{1-\bar{\alpha}z}}\) with \(|\alpha|<1\) and \({\mathbb D}=\{z:\ |z|<1\}\). Prove that
\(\displaystyle{\frac{1}{\pi}\iint_{{\mathbb D}} |\psi'_{\alpha}|^2 dx dy =1}\).
Show that \[ \frac{1}{\pi}\iint_{{\mathbb D}} |\psi'_{\alpha}| dx dy =\frac{1-|\alpha|^2}{|\alpha|^2}\log \frac{1}{1-|\alpha|^2}. \]
Prove that \(\displaystyle{f(z)=-\frac{1}{2}\left(z+\frac{1}{z}\right)}\) is a conformal map from half disc \(\{z=x+iy:\ |z|<1,\ y>0\}\) to upper half plane \(\mathbb H=\{z=x+iy:\ y>0\}\).
Solution
Goal:
For \(\psi_\alpha(z) = \frac{\alpha - z}{1 - \bar{\alpha}z}\) with \(|\alpha| < 1\), prove that \(\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_\alpha'(z)|^2 \,dx\,dy = 1\).
Prove that \(\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_\alpha'(z)| \,dx\,dy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\left(\frac{1}{1-|\alpha|^2}\right)\) (for \(\alpha \neq 0\), with limit \(1\) as \(\alpha \to 0\)).
Prove that \(f(z) = -\frac{1}{2}\left(z + \frac{1}{z}\right)\) is a conformal map from the upper half-disk \(\mathbb{D}^+ = \{z = x+iy : |z| < 1, y > 0\}\) to the upper half-plane \(\mathbb{H} = \{w = u+iv : v > 0\}\).
Part 1: Area Integral of \(|\psi_\alpha'(z)|^2\)
1. \(\psi_\alpha\) is a biholomorphic automorphism of the unit disk \(\mathbb{D}\).
1.1. \(\psi_\alpha\) is a Möbius transformation with pole at \(z = 1/\bar{\alpha}\).Since \(|\alpha| < 1\), \(|1/\bar{\alpha}| > 1\), so \(\psi_\alpha\) is holomorphic in a neighborhood of \(\overline{\mathbb{D}}\).Proof: Pole lies strictly outside the closed unit disk.
1.2. For \(|z| = 1\), \(|\psi_\alpha(z)|^2 = \frac{|\alpha - z|^2}{|1 - \bar{\alpha}z|^2} = \frac{(\alpha - z)(\bar{\alpha} - \bar{z})}{(1 - \bar{\alpha}z)(1 - \alpha \bar{z})} = \frac{|\alpha|^2 - \alpha \bar{z} - \bar{\alpha}z + 1}{1 - \alpha \bar{z} - \bar{\alpha}z + |\alpha|^2|z|^2} = 1\) (since \(|z|^2=1\)). Proof: Algebraic calculation on the boundary circle \(\partial \mathbb{D}\).
1.3. \(\psi_\alpha(\psi_\alpha(z)) = \frac{\alpha - \frac{\alpha-z}{1-\bar{\alpha}z}}{1 - \bar{\alpha}\frac{\alpha-z}{1-\bar{\alpha}z}} = \frac{\alpha(1-\bar{\alpha}z) - (\alpha-z)}{(1-\bar{\alpha}z) - \bar{\alpha}(\alpha-z)} = \frac{z(1-|\alpha|^2)}{1-|\alpha|^2} = z\).Proof: \(\psi_\alpha\) is its own inverse (an involution), hence a bijection of \(\mathbb{D}\) onto \(\mathbb{D}\).
1.4. Q.E.D.
2. The Jacobian determinant of a holomorphic mapping \(\psi(z) = u(x,y) + i v(x,y)\) is \(|\psi'(z)|^2\).
2.1. By the Cauchy-Riemann equations \(u_x = v_y\) and \(u_y = -v_x\), the Jacobian determinant is \(J_\psi(x,y) = u_x v_y - u_y v_x = u_x^2 + v_x^2 = |u_x + i v_x|^2 = |\psi'(z)|^2\).Proof: Standard real-differentiability computation for holomorphic functions.
2.2. Q.E.D.
3. Compute \(\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_\alpha'(z)|^2 \,dx\,dy\).
3.1. By the change of variables formula for 2D integrals under the diffeomorphism \(\psi_\alpha: \mathbb{D} \to \mathbb{D}\): \[\iint_{\mathbb{D}} |\psi_\alpha'(z)|^2 \,dx\,dy = \iint_{\mathbb{D}} J_{\psi_\alpha}(x,y) \,dx\,dy = \text{Area}(\psi_\alpha(\mathbb{D})) = \text{Area}(\mathbb{D}) = \pi.\] Proof: Change of variables theorem for planar integrals.
3.2. Dividing both sides by \(\pi\) yields \(\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_\alpha'(z)|^2 \,dx\,dy = \frac{\pi}{\pi} = 1\).Proof: Follows from
3.1.
3.3. Q.E.D.
Part 2: Area Integral of \(|\psi_\alpha'(z)|\)
4. Compute the derivative \(\psi_\alpha'(z)\).
4.1. By the quotient rule: \[\psi_\alpha'(z) = \frac{-(1-\bar{\alpha}z) - (\alpha-z)(-\bar{\alpha})}{(1-\bar{\alpha}z)^2} = \frac{-1 + \bar{\alpha}z + |\alpha|^2 - \bar{\alpha}z}{(1-\bar{\alpha}z)^2} = \frac{-(1-|\alpha|^2)}{(1-\bar{\alpha}z)^2}.\] Proof: Direct differentiation.
4.2. Therefore, \(|\psi_\alpha'(z)| = \frac{1-|\alpha|^2}{|1-\bar{\alpha}z|^2}\).Proof: Absolute value of
4.1.
4.3. Q.E.D.
5. Evaluate the integral in polar coordinates.
5.1. Without loss of generality, assume \(\alpha \neq 0\) (if \(\alpha = 0\), \(|\psi_0'(z)| = 1\), and \(\frac{1}{\pi}\iint_{\mathbb{D}} 1\,dx\,dy = 1 = \lim_{\alpha\to 0} \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}\)). Let \(\alpha = a e^{i\phi}\) with \(a = |\alpha| \in (0, 1)\).By rotational invariance of the disk, substituting \(z \mapsto e^{i\phi} z\) allows us to assume \(\alpha = a \in (0,1)\) is real: \[\iint_{\mathbb{D}} \frac{1-a^2}{|1-a z|^2} \,dx\,dy = (1-a^2) \int_0^1 r \left( \int_0^{2\pi} \frac{d\theta}{|1 - a r e^{i\theta}|^2} \right) dr.\] Proof: Polar coordinates \(z = r e^{i\theta}\) and rotation invariance of the area element \(dx\,dy = r\,dr\,d\theta\).
5.2. For any \(c \in (0, 1)\), \(\frac{1}{2\pi}\int_0^{2\pi} \frac{d\theta}{|1 - c e^{i\theta}|^2} = \frac{1}{1-c^2}\).Proof: Write \(\frac{1}{1-c e^{i\theta}} = \sum_{k=0}^\infty c^k e^{ik\theta}\).By Parseval’s identity / orthogonality of \(\{e^{ik\theta}\}_{k\in\mathbb{Z}}\), \(\frac{1}{2\pi}\int_0^{2\pi} \left|\sum_{k=0}^\infty c^k e^{ik\theta}\right|^2 d\theta = \sum_{k=0}^\infty (c^k)^2 = \sum_{k=0}^\infty c^{2k} = \frac{1}{1-c^2}\).
5.3. Setting \(c = ar\) with \(0 \leq ar < 1\), we get \(\int_0^{2\pi} \frac{d\theta}{|1 - a r e^{i\theta}|^2} = \frac{2\pi}{1 - a^2 r^2}\).Proof: Applied 2 with \(c = ar\).
5.4. Integrate with respect to \(r\): \[\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_a'(z)| \,dx\,dy = \frac{1-a^2}{\pi} \int_0^1 r \cdot \frac{2\pi}{1-a^2 r^2} \,dr = (1-a^2) \int_0^1 \frac{2r}{1-a^2 r^2} \,dr.\] Proof: Substitution of 3 into
5.1.
5.5. Using the substitution \(u = 1 - a^2 r^2\), \(du = -2a^2 r\,dr\): \[\int_0^1 \frac{2r}{1-a^2 r^2} \,dr = \int_{1-a^2}^1 \frac{du}{a^2 u} = \frac{1}{a^2} [\log u]_{1-a^2}^1 = \frac{1}{a^2} \log\left(\frac{1}{1-a^2}\right).\] Proof: Standard calculus integration.
5.6. Multiplying by \((1-a^2)\) gives: \[\frac{1}{\pi}\iint_{\mathbb{D}} |\psi_\alpha'(z)| \,dx\,dy = \frac{1-a^2}{a^2}\log\left(\frac{1}{1-a^2}\right) = \frac{1-|\alpha|^2}{|\alpha|^2}\log\left(\frac{1}{1-|\alpha|^2}\right).\] Proof: Combines 4 and
5.5.
5.7. Q.E.D.
Part 3: Conformal Map \(f(z) = -\frac{1}{2}\left(z + \frac{1}{z}\right)\) from \(\mathbb{D}^+\) to \(\mathbb{H}\)
6. \(f\) is holomorphic on \(\mathbb{D}^+\) with non-vanishing derivative.
6.1. \(f(z) = -\frac{1}{2}(z + z^{-1})\) is rational with its only pole at \(z=0 \notin \mathbb{D}^+\), hence holomorphic on \(\mathbb{D}^+\).Proof: \(0 \notin \mathbb{D}^+\) since points in \(\mathbb{D}^+\) have \(y = \text{Im}(z) > 0\).
6.2. \(f'(z) = -\frac{1}{2}\left(1 - \frac{1}{z^2}\right) = \frac{1 - z^2}{2z^2}\).Proof: Direct differentiation.
6.3. \(f'(z) = 0 \iff z^2 = 1 \iff z = \pm 1 \notin \mathbb{D}^+\).Thus \(f'(z) \neq 0\) for all \(z \in \mathbb{D}^+\).Proof: Points \(\pm 1\) have \(y=0\) and \(|z|=1\), so they lie on the boundary, not in \(\mathbb{D}^+\).
6.4. Q.E.D.
7. \(f\) maps \(\mathbb{D}^+\) into the upper half-plane \(\mathbb{H}\).
7.1. For \(z = r e^{i\theta} \in \mathbb{D}^+\) (so \(0 < r < 1\) and \(0 < \theta < \pi\)): \[f(z) = -\frac{1}{2}\left(r e^{i\theta} + \frac{1}{r} e^{-i\theta}\right) = -\frac{1}{2}\left( r(\cos\theta + i\sin\theta) + \frac{1}{r}(\cos\theta - i\sin\theta) \right).\] Proof: Euler’s formula.
7.2. The imaginary part is \(\text{Im}(f(z)) = -\frac{1}{2}\left(r - \frac{1}{r}\right)\sin\theta = \frac{1}{2}\left(\frac{1}{r} - r\right)\sin\theta\).Proof: Extracting the imaginary component.
7.3. Since \(0 < r < 1\), \(\frac{1}{r} - r > 0\); and since \(0 < \theta < \pi\), \(\sin\theta > 0\).Thus \(\text{Im}(f(z)) > 0\), so \(f(\mathbb{D}^+) \subseteq \mathbb{H}\).Proof: Product of two strictly positive real numbers is strictly positive.
7.4. Q.E.D.
8. \(f: \mathbb{D}^+ \to \mathbb{H}\) is bijective.
8.1. Let \(w \in \mathbb{H}\) be given.We solve \(f(z) = w \iff -\frac{1}{2}(z + 1/z) = w \iff z^2 + 2w z + 1 = 0\).Proof: Algebraic rearrangement.
8.2. The two roots of this quadratic equation are \(z_{1,2} = -w \pm \sqrt{w^2 - 1}\).The product of the roots is \(z_1 z_2 = 1\).Proof: Vieta’s formulas for \(z^2 + 2wz + 1 = 0\).
8.3. Since \(w \in \mathbb{H}\), \(w \notin [-1, 1]\), so \(w^2 - 1 \neq 0\) and the two roots are distinct and neither has modulus \(1\) (if \(|z|=1\) with \(z=e^{i\theta}\), then \(-\frac{1}{2}(e^{i\theta}+e^{-i\theta}) = -\cos\theta \in [-1,1] \subset \mathbb{R}\), which has imaginary part \(0\)). Proof: Modulus 1 points map to the real interval \([-1, 1]\).
8.4. Since \(|z_1||z_2| = 1\) and \(|z_1| \neq 1\), exactly one root has modulus strictly less than \(1\) (say \(|z_1| < 1\)). Proof: If \(|z_1| < 1\), then \(|z_2| = 1/|z_1| > 1\).
8.5. From <1>7.<2>2, \(\text{Im}(w) = \frac{1}{2}(\frac{1}{|z_1|} - |z_1|)\sin(\arg z_1)\).Since \(\text{Im}(w) > 0\) and \(\frac{1}{|z_1|} - |z_1| > 0\), we must have \(\sin(\arg z_1) > 0\), which means \(\arg(z_1) \in (0, \pi)\), so \(z_1 \in \mathbb{D}^+\).Proof: Sign of \(\sin(\arg z_1)\) must match the sign of \(\text{Im}(w)\).
8.6. Thus, for each \(w \in \mathbb{H}\), there is a unique \(z \in \mathbb{D}^+\) such that \(f(z) = w\).Proof: Exactly one of the two quadratic roots lies in \(\mathbb{D}^+\).
8.7. Q.E.D.
9. Conclusion: \(f\) is a conformal map (biholomorphism) from \(\mathbb{D}^+\) to \(\mathbb{H}\).
9.1. By 6 and 8, \(f\) is a holomorphic bijection from \(\mathbb{D}^+\) to \(\mathbb{H}\), hence a biholomorphism / conformal map.Proof: A bijective holomorphic function between open domains in \(\mathbb{C}\) has a holomorphic inverse.
9.2. Q.E.D.