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[parent] Möbius transformation cross-ratio preservation theorem (Theorem)

A Möbius transformation $f: z \mapsto w$ preserves the cross-ratios, i.e.

$\displaystyle \frac{(w_1-w_2)(w_3-w_4)}{(w_1-w_4)(w_3-w_2)} = \frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_3-z_2)} $

Conversely, given two quadruplets which have the same cross-ratio, there exists a Möbius transformation which maps one quadruplet to the other.

A consequence of this result is that the cross-ratio of $(a,b,c,d)$ is the value at $a$ of the Möbius transformation that takes $b$ $c$ $d$ to $1$ $0$ $\infty$ respectively.




"Möbius transformation cross-ratio preservation theorem" is owned by rspuzio. [ full author list (2) | owner history (1) ]
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See Also: cross ratio


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Proof of Möbius transformation cross-ratio preservation theorem (Proof) by Johan
proof of converse of Möbius transformation cross-ratio preservation theorem (Proof) by rspuzio
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Cross-references: consequence, maps, cross-ratio, quadruplets, conversely, preserves, Möbius transformation

This is version 6 of Möbius transformation cross-ratio preservation theorem, born on 2003-04-28, modified 2007-05-02.
Object id is 4222, canonical name is MobiusTransformationCrossRatioPreservationTheorem.
Accessed 4661 times total.

Classification:
AMS MSC30E20 (Functions of a complex variable :: Miscellaneous topics of analysis in the complex domain :: Integration, integrals of Cauchy type, integral representations of analytic functions)

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cross-ratio by Larry Hammick on 2003-08-02 01:39:30
Hi,
If it were me I would rename this item "cross-ratio", and include
1) the proposition that the cross-ratio (a,b,c,d) is the
value at a of the mobius transformation that takes b,c,d, to
1,0,infty respectively
2) a proof of the preservation, something like:
=====
Write
$$
g(z)=\frac{(z-z_2)(z_3-z_4)}{(z-z_4)(z_3-z_2)}\;.
$$
The function $gf^{-1}$ takes $f(z_2),f(z_3),f(z_4)$ to 1,0,$\infty$
respectively. So, by the above characterization of the cross ratio,
we have
$$g(f(z_1),f(z_2),f(z_3),f(z_4))=gf^{-1}(f(z_1))=g(z_1)\;.$$
=====
Larry
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