Condition for Points in Complex Plane to form Parallelogram
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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:
- $2 + i, 3 + 2 i, 2 + 3 i, 1 + 2 i$
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)