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Math Help - Cross-ratio

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Cross-ratio

    Prove that the cross-ratio of four complex numbers, defined as

    (z_1,z_2,z_3,z_4):=\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}

    is real if and only if the four points lie on a circle.
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    To clarify this problem

    I presume you mean the circumference and not inside of it.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    A circle is the boundary (circumference) of a disc.
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    Quote Originally Posted by Bruno J. View Post
    Prove that the cross-ratio of four complex numbers, defined as

    (z_1,z_2,z_3,z_4):=\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}


    is real if and only if the four points lie on a circle.


    Spoiler:
    Proof 1: as \arg\left(\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}\right)=\arg\left(\frac{z_1-z_3}{z_1-z_4}\right)-\arg\left(\frac{z_2-z_3}{z_2-z_2}\right) ( because \frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}=\frac{z_1-z_3}{z_1-z_4}:\frac{z_2-z_3}{z_2-z_4} ) , from basic geometry

    we get that z_i are on a circle iff the above is either the difference of two inscribed angles subtended on the same arc, and thus this difference is zero, or else

    this is the difference of two inscribed angles subtended over complementary arcs and thus the difference is \pm \pi (opposite angles of an quadrilateral inscribed

    in a circle sum 180 degrees).

    Proof 2: We use that the cross ratio is invariant under fractional transformations and that the cross ration is well defined if three points out of the four are distinct.

    Now, the frac. trans. T(z):=\frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)} maps z_1,\,z_2,\,z_3 to \infty,\,0,\,1 , resp., so z_1,z_2,z_3,z_4)" alt="T(z_4)=z_1,z_2,z_3,z_4)" /> is real iff z_1,z_2,z_3 lie on the circle (or on the

    straight line) on which z_1,z_2,z_3 lie

    Tonio
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Good job!

    Here's my solution:

    Let w(z) be given by \frac{(z-z_3)(z_2-z_4)}{(z_2-z_3)(z-z_4)} = \frac{(w-w_3)(w_2-w_4)}{(w_2-z_3)(w-w_4)}, where w_2, w_3, w_4 are any three distinct real points.

    Then z\mapsto w is a Möbius transformation, which takes circles to circles. The circle containing (z_2,z_3,z_4) is therefore mapped to the real line, since the three points z_2,z_3,z_4 are mapped to the real line.

    (This is essentially equivalent to your second solution!)
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    Quote Originally Posted by Bruno J. View Post
    Prove that the cross-ratio of four complex numbers, defined as

    (z_1,z_2,z_3,z_4):=\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}


    is real if and only if the four points lie on a circle.


    If the four complex numbers are real , your cross-ratio should be real ,so will we say that those four points lie on a circle with infinitely large radius ? ( Desargues's point of view )
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by simplependulum View Post
    If the four complex numbers are real , your cross-ratio should be real ,so will we say that those four points lie on a circle with infinitely large radius ? ( Desargues's point of view )
    Yes! A line is just a degenerate circle.
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