Cross-ratio

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June 8th 2010, 03:55 PM
Bruno J.

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.
June 8th 2010, 04:38 PM
wonderboy1953

To clarify this problem

I presume you mean the circumference and not inside of it.
June 8th 2010, 04:40 PM
Bruno J.

A circle is the boundary (circumference) of a disc.
June 8th 2010, 07:18 PM
tonio

Quote:

Originally Posted by Bruno J.

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 $T(z_4)=:(z_1,z_2,z_3,z_4)$ 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
June 8th 2010, 07:21 PM
Bruno J.

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!)
June 8th 2010, 08:47 PM
simplependulum

Quote:

Originally Posted by Bruno J.

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 )
June 8th 2010, 09:31 PM
Bruno J.

Quote:

Originally Posted by simplependulum

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.