# Cross-ratio

• Jun 8th 2010, 03:55 PM
Bruno J.
Cross-ratio
Prove that the cross-ratio of four complex numbers, defined as

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

is real if and only if the four points lie on a circle.
• Jun 8th 2010, 04:38 PM
wonderboy1953
To clarify this problem
I presume you mean the circumference and not inside of it.
• Jun 8th 2010, 04:40 PM
Bruno J.
A circle is the boundary (circumference) of a disc.
• Jun 8th 2010, 07:18 PM
tonio
Quote:

Originally Posted by Bruno J.
Prove that the cross-ratio of four complex numbers, defined as

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

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

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

we get that $\displaystyle z_i$ 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 $\displaystyle \pm \pi$ (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. $\displaystyle T(z):=\frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$ maps $\displaystyle z_1,\,z_2,\,z_3$ to $\displaystyle \infty,\,0,\,1$ , resp., so $\displaystyle T(z_4)=:(z_1,z_2,z_3,z_4)$ is real iff $\displaystyle z_1,z_2,z_3$ lie on the circle (or on the

straight line) on which $\displaystyle z_1,z_2,z_3$ lie

Tonio
• Jun 8th 2010, 07:21 PM
Bruno J.
Good job!

Here's my solution:

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

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

(This is essentially equivalent to your second solution!)
• Jun 8th 2010, 08:47 PM
simplependulum
Quote:

Originally Posted by Bruno J.
Prove that the cross-ratio of four complex numbers, defined as

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

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 )
• Jun 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.