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    Cross-ratio construction of conformal maps

    PR-AQ6YR

    Normalize both triples to $1,0,\infty$; composing the first normalization with the inverse of the second gives the Möbius transformation sending $z_1,z_2,z_3$ to $w_1,w_2,w_3$.Suggest a slogan

    For \(z_1\in\CC\) and distinct \(z_2,z_3,z_4\in\CC\), write \[ (z_1, z_2, z_3, z_4) \coloneqq {z_1 - z_3\over z_1-z_4}\cdot{z_2 - z_4 \over z_2 - z_3}. \]

    1. For distinct \(a,b,c\in\CC\), the map \[ R(z) \coloneqq (z, a,b,c) = {z - b\over z-c}\cdot{a - c\over a - b} \] is a Möbius transformation with \(R(a)=1\), \(R(b)=0\), and \(R(c)=\infty\).

    2. If one of \(a,b,c\) is \(\infty\), the factors containing it are deleted, and the resulting Möbius transformation has the same three values: \[ (z,\infty,b,c)={z-b\over z-c},\qquad (z,a,\infty,c)={a-c\over z-c},\qquad (z,a,b,\infty)={z-b\over a-b}. \]

    3. For distinct \(z_1,z_2,z_3\) and distinct \(w_1,w_2,w_3\) in \(\CC\cup\theset{\infty}\), let \(R_z(z)\coloneqq(z,z_1,z_2,z_3)\) and \(R_w(w)\coloneqq(w,w_1,w_2,w_3)\). Then \(T\coloneqq \inverseof{R_w}\circ R_z\) is a Möbius transformation with \(T(z_j)=w_j\) for \(j=1,2,3\).

    Parts (a) and (b) follow by substituting \(a\), \(b\), \(c\) into each formula; each formula is a quotient of affine functions of \(z\) with nonzero determinant because \(a,b,c\) are distinct. For (c), \(R_z\) sends \(z_1,z_2,z_3\) to \(1,0,\infty\), and \(\inverseof{R_w}\) sends \(1,0,\infty\) to \(w_1,w_2,w_3\); a composite of Möbius transformations is a Möbius transformation.

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