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Bilinear map preserves cross ratio

Short answer

For w=(az+b)/(cz+d), the difference formula gives w_i-w_j=(ad-bc)(z_i-z_j)/((cz_i+d)(cz_j+d)). Applying it to each cross-ratio factor cancels all denominators and (ad-bc)^2, proving the cross-ratio is invariant.

Define conformal mapping. Show that w = (az + b)/(cz + d) is invariant to [(w - w₁)/(w - w₃)] × [(w₂ - w₃)/(w₂ - w₁)] = [(z - z₁)/(z - z₃)] × [(z₂ - z₃)/(z₂ - z₁)].

A conformal mapping w = f(z) is a mapping that preserves angles between any two curves, both in magnitude and in sense (direction), at every point where f′(z) ≠ 0. Every analytic function with f′(z) ≠ 0 is conformal.

Claim

Let w = (az + b)/(cz + d) with ad − bc ≠ 0. Then the cross-ratio is unchanged: the expression in the w-plane equals the same expression in the z-plane.

Proof

For any two points zᵢ, zⱼ with images wᵢ, wⱼ:

wᵢ − wⱼ = (azᵢ + b)/(czᵢ + d) − (azⱼ + b)/(czⱼ + d) = [(azᵢ + b)(czⱼ + d) − (azⱼ + b)(czᵢ + d)] / [(czᵢ + d)(czⱼ + d)] = (ad − bc)(zᵢ − zⱼ) / [(czᵢ + d)(czⱼ + d)]. ... (1)

(The same formula holds for any variable point z with image w.) Now apply (1) to each factor, writing Dᵢ = czᵢ + d and D = cz + d:

(w − w₁)(w₂ − w₃)
 = (ad−bc)² (z − z₁)(z₂ − z₃) / [ D·D₁ · D₂·D₃ ]

(w − w₃)(w₂ − w₁)
 = (ad−bc)² (z − z₃)(z₂ − z₁) / [ D·D₃ · D₂·D₁ ]

Both products have the same denominator D·D₁·D₂·D₃ and the same factor (ad − bc)², so they cancel on division:

(w − w₁)(w₂ − w₃) / [(w − w₃)(w₂ − w₁)] = (z − z₁)(z₂ − z₃) / [(z − z₃)(z₂ − z₁)]. ∎

So the cross-ratio is invariant under a bilinear transformation. This is why the relation can be used to find the transformation that maps three given points z₁, z₂, z₃ to w₁, w₂, w₃.

✓ Answer checked by Claude Opus 5.5 (high effort)

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