Quadrilateral in Complex Plane is Cyclic iff Cross Ratio of Vertices is Real

Theorem
Let $z_1, z_2, z_3, z_4$ be distinct complex numbers.

Then:
 * $z_1, z_2, z_3, z_4$ define the vertices of a cyclic quadrilateral

their cross-ratio:
 * $\paren {z_1, z_3; z_2, z_4} = \dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} }$

is wholly real.

Proof
Let $z_1 z_2 z_3 z_4$ be a cyclic quadrilateral.

By Geometrical Interpretation of Complex Subtraction, the four sides of $z_1 z_2 z_3 z_4$ can be defined as $z_1 - z_2$, $z_3 - z_2$, $z_3 - z_4$ and $z_1 - z_4$.


 * Cyclic-Quadrilateral-in-Complex-Plane.png

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, the opposite angles of $z_1 z_2 z_3 z_4$ sum to $\pi$ radians.

By Complex Multiplication as Geometrical Transformation, it follows that:

Thus:

Hence the result.