Chapter 8 • Theory & Derivations

Conformal Mappings, Möbius Transformations & Analytic Continuation

Angle and scale preservation of conformal maps, Möbius bilinear transformations, cross-ratio invariance, circle-to-circle property, Joukowsky airfoil maps, and analytic continuation.

§8.1 Conformal Mapping Theory, Angle Preservation & Scale Invariance

1. Definition of Conformal Mapping

A mapping $w = f(z)$ is called conformal at $z_0$ if it preserves both the magnitude and sense (orientation) of angles between any two smooth intersecting curves at $z_0$.

Theorem 8.1 (Conformality Criterion): Let $f(z)$ be holomorphic in a domain $D$. Then $f(z)$ is conformal at every point $z_0 \in D$ where $f'(z_0) \ne 0$.

Points where $f'(z_0) = 0$ are called critical points. At a critical point where $f^{(k)}(z_0) = 0$ for $k = 1, \dots, m-1$ and $f^{(m)}(z_0) \ne 0$, angles between curves are multiplied by factor $m$.

§8.2 Bilinear (Möbius) Transformations w = (az + b)/(cz + d) & Cross-Ratio Invariance

1. Möbius (Bilinear) Transformations

A Möbius transformation is a rational function of the form:

$$w = T(z) = \frac{az + b}{cz + d}, \quad a, b, c, d \in \mathbb{C}, \quad ad - bc \ne 0$$

Under functional composition, Möbius transformations form the projective linear group $\text{PGL}(2, \mathbb{C})$ acting as conformal automorphisms of the extended complex plane $\widehat{\mathbb{C}}$.

2. The Circle-to-Circle Property

Theorem 8.2 (Preservation of Generalized Circles): Every Möbius transformation maps generalized circles (which include straight lines as circles of infinite radius) to generalized circles.

3. Cross-Ratio Invariance

Theorem 8.3 (Cross-Ratio Invariance): For any four distinct points $z_1, z_2, z_3, z_4 \in \widehat{\mathbb{C}}$, the cross-ratio: $$(z_1, z_2; z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}$$ is strictly invariant under any Möbius transformation $T$: $$(T(z_1), T(z_2); T(z_3), T(z_4)) = (z_1, z_2; z_3, z_4)$$ Consequently, the unique Möbius transformation mapping three given points $z_1, z_2, z_3$ to $w_1, w_2, w_3$ is determined by equating $(w, w_1; w_2, w_3) = (z, z_1; z_2, z_3)$.

§8.3 Canonical Mappings: Upper Half-Plane to Unit Disk & Joukowsky Transformation

1. The Cayley Transform: Upper Half-Plane to Unit Disk

The upper half-plane $\mathbb{H} = \{\text{Im}(z) > 0\}$ is conformally mapped onto the open unit disk $\mathbb{D} = \{|w| < 1\}$ by the Cayley transform:

$$w = e^{i\theta} \frac{z - z_0}{z - \bar{z}_0}, \quad z_0 \in \mathbb{H}, \quad \theta \in \mathbb{R}$$

The real axis $\mathbb{R} = \partial \mathbb{H}$ maps onto the unit circle $\partial \mathbb{D} = \{|w| = 1\}$.

2. The Joukowsky Airfoil Transformation

The Joukowsky transformation maps fluid flow past a circular cylinder onto flow past an aerodynamic airfoil with a sharp trailing edge:

$$w = z + \frac{c^2}{z}, \quad c > 0$$

Critical points occur where $\frac{dw}{dz} = 1 - \frac{c^2}{z^2} = 0 \implies z = \pm c$. The circle $|z| = c$ collapses onto the real segment $[-2c, 2c]$. A circle passing through $z = -c$ and containing $z = c$ maps onto a streamlined airfoil shape (Joukowsky profile).

§8.4 Analytic Continuation, Schwarz Reflection Principle & Riemann Surfaces

1. Analytic Continuation and Permanence of Functional Relations

If two holomorphic functions $f_1, f_2$ on domains $D_1, D_2$ agree on a non-empty open overlap $D_1 \cap D_2$, then $f_2$ is the unique analytic continuation of $f_1$ into $D_2$.

2. The Schwarz Reflection Principle

Theorem 8.4 (Schwarz Reflection Principle): Let $D$ be a domain symmetric with respect to the real axis ($z \in D \iff \bar{z} \in D$), and let $D^+ = \{z \in D : \text{Im}(z) > 0\}$. If $f(z)$ is holomorphic in $D^+$, continuous on $D^+ \cup (D \cap \mathbb{R})$, and real-valued on $D \cap \mathbb{R}$, then $f$ can be analytically continued to the entire domain $D$ by defining: $$f(z) = \overline{f(\bar{z})} \quad \text{for } z \in D^-$$

3. Riemann Surfaces

To eliminate branch cuts and render multivalued functions globally single-valued and holomorphic, Bernhard Riemann introduced Riemann surfaces: multi-sheeted branched covering spaces over $\mathbb{C}$. For example, $w = \sqrt{z}$ is a single-valued holomorphic function on a two-sheeted helical Riemann surface connected along the branch cut.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Example 8.1: Möbius Transformation via Cross-Ratio

Find the unique Möbius transformation $w = T(z)$ mapping $z_1 = 0, z_2 = 1, z_3 = \infty$ to $w_1 = -1, w_2 = -i, w_3 = 1$.

Step 1: Cross-ratio formula
The cross-ratio $(w, w_1; w_2, w_3) = (z, z_1; z_2, z_3)$ is invariant:

\frac{(w - w_2)(w_1 - w_3)}{(w - w_3)(w_1 - w_2)} = \frac{(z - z_2)(z_1 - z_3)}{(z - z_3)(z_1 - z_2)}

Step 2: Handle points with $\infty$
Since $z_3 = \infty$, the RHS simplifies to:

\frac{z - z_2}{z_1 - z_2} = \frac{z - 1}{0 - 1} = -(z - 1) = 1 - z

Step 3: Substitute $w$-points into LHS
$w_1 = -1, w_2 = -i, w_3 = 1$:

\frac{(w - (-i))(-1 - 1)}{(w - 1)(-1 - (-i))} = \frac{(w + i)(-2)}{(w - 1)(-1 + i)} = \frac{2(w + i)}{(1 - i)(w - 1)}

Multiply numerator and denominator by $1 + i$: $\frac{2}{1 - i} = \frac{2(1 + i)}{2} = 1 + i$.

(1 + i) \frac{w + i}{w - 1} = 1 - z

Step 4: Solve for $w$

(1 + i)(w + i) = (1 - z)(w - 1) \implies (1 + i)w + i - 1 = w(1 - z) - (1 - z)

$$w[(1 + i) - (1 - z)] = -(1 - z) - (i - 1) = z - 1 - i + 1 = z - i$$

w(z + i) = z - i \implies w = \frac{z - i}{z + i}

Verify: $z = 0 \implies w = -i/i = -1$. $z = 1 \implies w = \frac{1 - i}{1 + i} = -i$. $z \to \infty \implies w \to 1$. Perfect!

Final Answer & Physical Insight

$w = \frac{z - i}{z + i}$

Tier 2 • Intermediate Exam Example 8.2: Conformal Exponential Mapping of Infinite Horizontal Strip

Determine the conformal image of the horizontal strip $S = \{z = x + iy : -\infty < x < \infty, 0 < y < \pi\}$ under $w = e^z$.

Step 1: Coordinate transformation
Let $z = x + iy \implies w = u + iv = e^{x + iy} = e^x (\cos y + i\sin y)$. In polar form $w = \rho e^{i\phi}$:

\rho = |w| = e^x, \qquad \phi = \arg w = y

Step 2: Range of modulus and argument

As $x \in (-\infty, \infty)$, the modulus $\rho = e^x$ ranges over $(0, \infty)$.
As $y \in (0, \pi)$, the argument $\phi = y$ ranges over $(0, \pi)$.

Step 3: Boundary mapping

Lower boundary $y = 0$: $w = e^x (\cos 0 + i\sin 0) = e^x > 0$. Maps to the positive real axis $(0, \infty)$.
Upper boundary $y = \pi$: $w = e^x (\cos\pi + i\sin\pi) = -e^x < 0$. Maps to the negative real axis $(-\infty, 0)$.
As $x \to -\infty$, $w \to 0$ (the origin).

Conclusion: The infinite horizontal strip $0 < y < \pi$ is mapped conformally and bijectively onto the Upper Half-Plane $\mathbb{H} = \{w \in \mathbb{C} : \text{Im}(w) > 0\}$.

Final Answer & Physical Insight

Upper Half-Plane $\mathbb{H} = \{w \in \mathbb{C} : \text{Im}(w) > 0\}$.

Tier 3 • Honors Challenge Example 8.3: Joukowsky Transformation & Trailing Edge Singularity Analysis

For the Joukowsky map $w = z + \frac{c^2}{z}$ ($c > 0$), prove that a circle $C$ passing through $z = -c$ and containing $z = c$ is mapped onto a streamlined airfoil profile with a cusped trailing edge of interior angle zero.

Step 1: Derivative and critical points

\frac{dw}{dz} = 1 - \frac{c^2}{z^2} = \frac{z^2 - c^2}{z^2} = \frac{(z - c)(z + c)}{z^2}

The critical points where $w'(z) = 0$ are $z = c$ and $z = -c$.

Step 2: Circle through $z = -c$
Let $C$ be a circle centered at $z_0 = -\mu + i\eta$ ($\mu > 0, \eta > 0$) with radius $R = |z_0 - (-c)|$ so that $z = -c$ lies on $C$, and $z = c$ lies strictly inside $C$. At $z = -c$, $w(-c) = -c - c = -2c$, which forms the trailing edge of the airfoil.

Step 3: Expansion near the critical point $z = -c$
Let $z = -c + \Delta z$.

w(z) = (-c + \Delta z) + \frac{c^2}{-c + \Delta z} = -c + \Delta z - c \left(1 + \frac{\Delta z}{c} + \frac{(\Delta z)^2}{c^2} + \dots\right)

= -2c + \frac{(\Delta z)^2}{c} + O((\Delta z)^3)

Notice that the linear term $\Delta z$ is absent!

w - (-2c) \approx \frac{1}{c} (\Delta z)^2

Step 4: Angle doubling and cusped trailing edge
Taking the argument:

\arg(w + 2c) \approx 2 \arg(\Delta z)

As the circle $C$ passes smoothly through $z = -c$, the tangent vector has angle difference $\pi$ between arriving and departing branches. Multiplying by 2 maps the angle $\pi$ to $2\pi$! This wraps the upper and lower surfaces together tangentially, creating a sharp cusp (interior angle $0^\circ$) at the trailing edge $w = -2c$. This geometric property satisfies the physical Kutta condition of fluid circulation!

Final Answer & Physical Insight

Proved: $w + 2c \approx \frac{(\Delta z)^2}{c}$ doubles the angle, forming a cusp at trailing edge $w = -2c$.