Conformal Mapping, Bilinear (Mobius) Transformations & Cross-Ratio Invariance

Explanation

Conformal mapping allows complicated physical flow and potential problems (such as aerodynamic flow around an airplane airfoil) to be mapped into simple geometric boundaries (such as a circular cylinder) where solutions are classical, and then transformed back without altering orthogonal field boundaries.

Key points

• Condition for Conformality: f(z) must be analytic at z₀ and f'(z₀) ≠ 0. Points where f'(z) = 0 are called Critical Points.
• Bilinear (Mobius) Transformation: w = (az + b) / (cz + d) where a, b, c, d are complex constants and ad - bc ≠ 0 (determinant non-zero).
• Decomposition: Every bilinear transformation is a composition of Translation (w = z + α), Rotation & Magnification (w = β z), and Inversion (w = 1/z).
• Circle-Preserving Property: Bilinear transformations map generalized circles (circles and straight lines) in the z-plane into generalized circles in the w-plane.
• Cross-Ratio Invariance: The cross-ratio of four distinct points is invariant under bilinear transformation:
• [(w - w₁)(w₂ - w₃)] / [(w - w₃)(w₂ - w₁)] = [(z - z₁)(z₂ - z₃)] / [(z - z₃)(z₂ - z₁)].
• Fixed Points (Invariant Points): Roots of z = (az + b)/(cz + d) => c z² + (d - a) z - b = 0 (at most 2 fixed points).

Example

Code examples

// Finding Bilinear Transformation via Cross-Ratio Invariance:
// Problem: Find the bilinear transformation that maps:
//    z1 = 1,  z2 = i,  z3 = -1   into
//    w1 = i,  w2 = 0,  w3 = -i
// Step 1: Set up Cross-Ratio Equation:
//    (w - w1)(w2 - w3) / [ (w - w3)(w2 - w1) ] = (z - z1)(z2 - z3) / [ (z - z3)(z2 - z1) ]
// Step 2: Substitute values:
//    LHS = (w - i)(0 - (-i)) / [ (w - (-i))(0 - i) ]
//        = (w - i)(i) / [ (w + i)(-i) ] = -(w - i) / (w + i) = (i - w) / (i + w)
//    RHS = (z - 1)(i - (-1)) / [ (z - (-1))(i - 1) ]
//        = (z - 1)(1 + i) / [ (z + 1)(i - 1) ]
//    Notice: (1 + i) / (i - 1) = (1 + i)^2 / (i^2 - 1) = (1 + 2i - 1) / (-2) = -i
//    RHS = -i (z - 1) / (z + 1)
// Step 3: Equate LHS and RHS:
//    -(w - i) / (w + i) = -i (z - 1) / (z + 1)
//    (w - i) / (w + i) = i (z - 1) / (z + 1)
// Step 4: Cross-multiply and solve for w:
//    (w - i)(z + 1) = i (z - 1)(w + i) = (i z - i)(w + i)
//    wz + w - iz - i = i z w - z + w + 1
//    wz + w - i z w - w = iz + i - z + 1
//    w [ z (1 - i) ] = z (i - 1) + (1 + i)
//    w = [ (i - 1) z + (1 + i) ] / [ (1 - i) z ]
//    Divide numerator and denominator by (1 - i):
//    w = -z / z + (1 + i)/(1 - i) / z => w = (i - z) / (i z + 1)

Exam questions

• [8 Marks] Find the bilinear transformation which maps z = {2, 1, 0} into w = {1, 0, i}.
• [7 Marks] Show that the transformation w = 1/z maps circles and straight lines in the z-plane into circles or straight lines in the w-plane.
• [7 Marks] Find the critical points of the transformation w = z + 1/z.