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.

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- Jun 8th 2010, 03:55 PMBruno 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 PMwonderboy1953To clarify this problem
I presume you mean the circumference and not inside of it.

- Jun 8th 2010, 04:40 PMBruno J.
A circle is the boundary (circumference) of a disc.

- Jun 8th 2010, 07:18 PMtonio
- Jun 8th 2010, 07:21 PMBruno 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 PMsimplependulum
- Jun 8th 2010, 09:31 PMBruno J.