Condition for Points in Complex Plane to form Parallelogram/Examples/2+i, 3+2i, 2+3i, 1+2i
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Theorem
The points in the complex plane represented by the complex numbers:
- $2 + i, 3 + 2 i, 2 + 3 i, 1 + 2 i$
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 $2$.