Condition for Points in Complex Plane to form Parallelogram

Theorem

Let $A = z_1$, $B = z_2$, $C = z_3$ and $D = z_4$ represent on the complex plane the vertices of a quadrilateral.

Then $ABCD$ is a parallelogram if and only if:

$z_1 - z_2 + z_3 - z_4 = 0$

Proof

$ABCD$ is a parallelogram if and only if:

$\vec {AB} = \vec {DC}$

Then:

\(\ds \vec {AB}\)

\(=\)

\(\ds \vec {DC}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds z_2 - z_1\)

\(=\)

\(\ds z_3 - z_4\)

\(\ds \leadstoandfrom \ \ \)

\(\ds z_1 - z_2 + z_3 - z_4\)

\(=\)

\(\ds 0\)

$\blacksquare$

Examples

$2 + i$, $3 + 2 i$, $2 + 3 i$, $1 + 2 i$ form Square

The points in the complex plane represented by the complex numbers:

$2 + i, 3 + 2 i, 2 + 3 i, 1 + 2 i$

are the vertices of a square.

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $65$

(although see Condition for Points in Complex Plane to form Parallelogram: Mistake for analysis of an error in that work)