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Conformal Mappings | Mathematics

Definition. An analytic function ff is conformal at z0z_0 if f"(z0)0f"(z_0) \neq 0. A conformal Mapping preserves angles (both magnitude and orientation) between curves.

If f(z0)=reiθf'(z_0) = re^{i\theta} Then near z0z_0 the mapping ff acts as a rotation by θ\theta followed By a scaling by rr. The Jacobian determinant is f(z0)2>0|f'(z_0)|^2 \gt 0 So orientation is preserved.

MappingEffect
w=az+bw = az + b (a0a \neq 0)Translation, rotation, scaling
w=1/zw = 1/zInversion in the unit circle
w=z2w = z^2Squaring (doubles angles)
w=ezw = e^zExponential (maps strips to sectors)
w=za1aˉzw = \frac{z - a}{1 - \bar{a}z}Möbius (maps disk to disk)

A Möbius transformation (or linear fractional transformation) is

T(z)=az+bcz+d,adbc0T(z) = \frac{az + b}{cz + d}, \quad ad - bc \neq 0

Proposition 10.1. Möbius transformations are conformal (where defined) and map circles and lines To circles and lines.

Proposition 10.2. Three points determine a unique Möbius transformation: T(z1)=w1T(z_1) = w_1 T(z2)=w2T(z_2) = w_2, T(z3)=w3T(z_3) = w_3.

Definition. The cross-ratio of four distinct points z1,z2,z3,z4z_1, z_2, z_3, z_4 is

(z1,z2,z3,z4)=(z1z3)(z2z4)(z1z4)(z2z3)(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)}

Proposition 10.3. The cross-ratio is invariant under Möbius transformations: (Tz1,Tz2,Tz3,Tz4)=(z1,z2,z3,z4)(Tz_1, Tz_2, Tz_3, Tz_4) = (z_1, z_2, z_3, z_4).

Proposition 10.4. The unique Möbius transformation sending z10z_1 \mapsto 0, z21z_2 \mapsto 1 z3z_3 \mapsto \infty is

T(z)=(zz1)(z2z3)(zz3)(z2z1)T(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}

10.6 Classification of Möbius Transformations

Section titled “10.6 Classification of Möbius Transformations”

A Möbius transformation T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} is classified by its fixed points (solutions of T(z)=zT(z) = z).

  1. Parabolic: Exactly one fixed point. Conjugate to w=z+kw = z + k.
  2. Elliptic: Two fixed points, T(z0)=1|T'(z_0)| = 1. Conjugate to a rotation w=eiθzw = e^{i\theta} z.
  3. Hyperbolic: Two fixed points, T(z0)R+T'(z_0) \in \mathbb{R}^+, T(z0)1T'(z_0) \neq 1. Conjugate to w=kzw = kz.
  4. Loxodromic: Two fixed points, T(z0)R{z:z=1}T'(z_0) \notin \mathbb{R} \cup \{z : |z| = 1\}. Conjugate to w=keiθzw = ke^{i\theta}z.
Solution

Problem. Find the Möbius transformation mapping 0i0 \mapsto i, 101 \mapsto 0, i\infty \mapsto -i.

T(z)=az+bcz+dT(z) = \frac{az + b}{cz + d} with T(0)=ib/d=ib=idT(0) = i \Rightarrow b/d = i \Rightarrow b = id. T(1)=0a=b=idT(1) = 0 \Rightarrow a = -b = -id. T()=ia/c=ic=dT(\infty) = -i \Rightarrow a/c = -i \Rightarrow c = d.

T(z)=idz+iddz+d=i(1z)z+1T(z) = \frac{-idz + id}{dz + d} = \frac{i(1 - z)}{z + 1}.

Problem. Show that T(z)=z1z+1T(z) = \frac{z - 1}{z + 1} maps the right half-plane to the unit disk.

If Re(z)>0\mathrm{Re}(z) \gt 0 Then z1<z+1|z - 1| \lt |z + 1| So T(z)<1|T(z)| \lt 1.

Check boundary: T(i)=i1i+1=(i1)(i+1)(i+1)(i+1)=22=1T(i) = \frac{i - 1}{i + 1} = \frac{(i-1)(-i+1)}{(i+1)(-i+1)} = \frac{2}{2} = 1. T(i)=1|T(i)| = 1. \checkmark

Problem. Classify T(z)=2z+1z+2T(z) = \frac{2z + 1}{z + 2}.

Fixed points: z=2z+1z+2z2=1z=±1z = \frac{2z + 1}{z + 2} \Rightarrow z^2 = 1 \Rightarrow z = \pm 1.

T(z)=3(z+2)2T'(z) = \frac{3}{(z + 2)^2}. T(1)=1/3T'(1) = 1/3, T(1)=3T'(-1) = 3.

Both multipliers are real and positive (not equal to 11), so TT is hyperbolic.

Theorem 10.5 (Riemann Mapping Theorem). Let UU be a connected open proper subset of C\mathbb{C}. Then there exists a bijective conformal map from UU onto the unit disk D={z:z<1}\mathbb{D} = \{z : |z| \lt 1\}.

This is one of the most profound results in complex analysis, establishing that all connected Domains (other than C\mathbb{C} itself) are conformally equivalent.

Remark. The Riemann mapping theorem is an existence theorem; it does not provide an explicit Formula for the conformal map .

Fluid dynamics. The complex potential w=f(z)=ϕ+iψw = f(z) = \phi + i\psi for a 2D incompressible, irrotational flow satisfies Laplace’s equation. Conformal mappings transform simple flow patterns (e.g., uniform flow past a circle) into flows past arbitrary smooth boundaries. The Joukowski transform w=z+1/zw = z + 1/z maps a circle to an airfoil shape, enabling analytical calculation of lift.

Electrostatics. The electric potential in a charge-free region satisfies 2V=0\nabla^2 V = 0. Conformal mappings transform the boundary value problem into a simpler geometry (e.g., upper half-plane or unit disk) where the solution is known, then map the solution back.

Heat conduction. Steady-state temperature distributions satisfy Laplace’s equation. Conformal mappings solve heat flow problems in irregularly shaped regions by mapping to canonical domains.

Key insight: Any problem governed by Laplace’s equation in 2D can be solved by conformally mapping the domain to a half-plane or disk, solving there, and mapping back.

Problem. Use the Joukowski transform to find the complex potential for flow past a cylinder.

Solution. The complex potential for uniform flow past a circle of radius RR centered at the origin is:

w(ζ)=U(ζ+R2ζ)w(\zeta) = U\left(\zeta + \frac{R^2}{\zeta}\right)

The Joukowski transform z=ζ+1ζz = \zeta + \frac{1}{\zeta} maps the circle ζ=R|\zeta| = R to an ellipse (or airfoil for RR near 1 with slight offset). Substituting ζ\zeta as a function of zz and composing gives the flow past the transformed body.

The velocity components are obtained from:

vxivy=dwdz=dwdζdζdzv_x - iv_y = \frac{dw}{dz} = \frac{dw}{d\zeta}\cdot\frac{d\zeta}{dz}

At infinity, vx=Uv_x = U and vy=0v_y = 0 (uniform flow). On the cylinder surface, the flow is tangent to the boundary (no penetration condition). \blacksquare

PropertyStatement
Angle preservationConformal maps preserve angles between intersecting curves
Local linearisationNear z0z_0, ff acts as rotation by argf(z0)\arg f'(z_0) and scaling by $
Circle preservationMöbius transformations map circles and lines to circles and lines
Cross-ratio invariance(Tz1,Tz2,Tz3,Tz4)=(z1,z2,z3,z4)(Tz_1, Tz_2, Tz_3, Tz_4) = (z_1, z_2, z_3, z_4) for any Möbius TT
Riemann mappingAny directly connected domain (≠ C\mathbb{C}) is conformally equivalent to D\mathbb{D}
Laplace correspondenceSolutions to 2u=0\nabla^2 u = 0 are preserved under conformal maps
flowchart TD
A[10_Conformal Mappings] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Conformal mappings are angle-preserving transformations of the complex plane. A holomorphic function with non-zero derivative acts locally as a rotation plus a scaling — it preserves the shape of infinitesimal figures while possibly changing their size and orientation. This makes conformal maps the natural language for problems involving fluid flow, electrostatics, and heat conduction, where the geometry of the domain can be simplified by mapping it to a simpler shape. Möbius transformations are the building blocks: they map circles to circles and are determined by where they send three points. The Riemann mapping theorem guarantees that any directly connected domain (except the whole plane) can be conformally mapped to the unit disk.

Mistake 1: Assuming that all analytic functions are conformal. An analytic function is conformal only where its derivative is non-zero. At points where f(z0)=0f'(z_0) = 0, the mapping is not conformal (angles are not preserved). For example, f(z)=z2f(z) = z^2 is conformal everywhere except at z=0z = 0, where it doubles angles.

Mistake 2: Confusing conformal with bijective. A conformal map need not be bijective. For example, f(z)=z2f(z) = z^2 is conformal on C{0}\mathbb{C} \setminus \{0\} but not injective (both zz and z-z map to the same point). A conformal bijection is called a biholomorphism or conformal equivalence.

Mistake 3: Forgetting that Möbius transformations map circles and lines to circles and lines. Möbius transformations map circles and lines to circles and lines, but they do not necessarily map a circle to a circle and a line to a line. A circle can be mapped to a line (if the circle passes through the pole of the transformation) and vice versa.

Mistake 4: Assuming that conformal maps preserve distances. Conformal maps preserve angles but not distances. A conformal map can stretch or compress regions while preserving angles. For example, f(z)=2zf(z) = 2z doubles all distances but preserves angles.

Mistake 5: Forgetting the cross-ratio invariance property. The cross-ratio (z1,z2,z3,z4)(z_1, z_2, z_3, z_4) is invariant under Möbius transformations. This property is useful for constructing Möbius transformations that map three given points to three specified points. Do not forget to use the cross-ratio when solving such problems.