Condition for Points in Complex Plane to form Parallelogram/Examples/2+i, 3+2i, 2+3i, 1+2i

Theorem

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.

Proof

Let us label the points:

\(\ds A\)

\(:=\)

\(\ds 2 + i\)

\(\ds B\)

\(:=\)

\(\ds 3 + 2 i\)

\(\ds C\)

\(:=\)

\(\ds 2 + 3 i\)

\(\ds D\)

\(:=\)

\(\ds 1 + 2 i\)

From Geometrical Interpretation of Complex Subtraction, we have that the difference $p - q$ between two complex numbers $p, q$ represents the vector $\vec {q p}$.

Let us take the differences of the complex numbers given:

\(\ds B - A: \ \ \)

\(\ds \paren{3 + 2 i} - \paren{2 + i}\)

\(=\)

\(\ds 1 + i\)

\(\ds B - C: \ \ \)

\(\ds \paren{3 + 2 i} - \paren{2 + 3 i}\)

\(=\)

\(\ds 1 - i\)

\(\ds B - D: \ \ \)

\(\ds \paren{3 + 2 i} - \paren{1 + 2 i}\)

\(=\)

\(\ds 2 + 0 i\)

\(\ds C - A: \ \ \)

\(\ds \paren{2 + 3 i} - \paren{2 + i}\)

\(=\)

\(\ds 0 + 2 i\)

\(\ds C - D: \ \ \)

\(\ds \paren{2 + 3 i} - \paren{1 + 2 i}\)

\(=\)

\(\ds 1 + i\)

\(\ds A - D: \ \ \)

\(\ds \paren{2 + i} - \paren{1 + 2 i}\)

\(=\)

\(\ds 1 - i\)

So:

$\vec {AB} = \vec {DC} = 1 + i$

$\vec {DA} = \vec {CB} = 1 - i$

Thus by definition $ABCD$ forms a parallelogram.

Next it is noted that:

$\paren {1 + i} i = i + i^2 = 1 - i$

and so $AB$ and $DC$ are perpendicular to $DA$ and $CB$.

Thus by definition $ABCD$ is a rectangle.

Finally note that:

$\cmod {1 + i} = \cmod {1 - i} = \sqrt {1^2 + 1^2} = \sqrt 2$

and so:

$\cmod {AB} = \cmod {DC} = \cmod {DC} = \cmod {DC}$

That is, all four sides of $ABCD$ are the same length.

Thus by definition $ABCD$ is a square.

$\blacksquare$

Sources

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