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

if and only if 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 $z_1 - z_4$.

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:

\(\ds \angle \, z_1\)

\(=\)

\(\ds \map \arg {\dfrac {z_1 - z_2} {z_1 - z_4} }\)

\(\ds \angle \, z_2\)

\(=\)

\(\ds \map \arg {\dfrac {z_3 - z_2} {z_1 - z_2} }\)

\(\ds \angle \, z_3\)

\(=\)

\(\ds \map \arg {\dfrac {z_3 - z_4} {z_3 - z_2} }\)

\(\ds \angle \, z_4\)

\(=\)

\(\ds \map \arg {\dfrac {z_1 - z_4} {z_3 - z_4} }\)

Thus:

\(\ds \angle \, z_1 + \angle \, z_3\)

\(=\)

\(\ds \pi\)

\(\ds \leadsto \ \ \)

\(\ds \map \arg {\dfrac {z_1 - z_2} {z_1 - z_4} } + \arg \paren {\dfrac {z_3 - z_4} {z_3 - z_2} }\)

\(=\)

\(\ds \pi\)

\(\ds \leadsto \ \ \)

\(\ds \map \arg {\paren {\dfrac {z_1 - z_2} {z_1 - z_4} } \paren {\dfrac {z_3 - z_4} {z_3 - z_2} } }\)

\(=\)

\(\ds \pi\)

Argument of Product equals Sum of Arguments

\(\ds \leadsto \ \ \)

\(\ds \map \Im {\dfrac {\paren {z_1 - z_2} \paren {z_3 - z_4} } {\paren {z_1 - z_4} \paren {z_3 - z_2} } }\)

\(=\)

\(\ds 0\)

Argument of Negative Real Number is Pi

Hence the result.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $6$.