Equation relating Points of Parallelogram in Complex Plane

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Theorem

Let $ABVU$ be a parallelogram in the complex plane whose vertices correspond to the complex numbers $a, b, v, u$ respectively.

Let $\angle BAU = \alpha$.

Let $\cmod {UA} = \lambda \cmod {AB}$.

Parallelogram-in-Complex-Plane.png

Then:

$u = \paren {1 - q} a + q b$
$v = -q a + \paren {1 + q} b$

where:

$q = \lambda e^{i \alpha}$


Proof

From Geometrical Interpretation of Complex Subtraction, the four sides of $UABC$ can be defined as:

\(\ds UA\) \(=\) \(\ds a - u\)
\(\ds \leadsto \ \ \) \(\ds AU\) \(=\) \(\ds u - a\) (easier in this form)
\(\ds AB\) \(=\) \(\ds b - a\)
\(\ds UV\) \(=\) \(\ds v - u\)
\(\ds BV\) \(=\) \(\ds v - b\)


Thus:

\(\ds UA\) \(=\) \(\ds \lambda \paren {AB}\)
\(\ds \leadsto \ \ \) \(\ds \cmod {a - u}\) \(=\) \(\ds \lambda \cmod {b - a}\)
\(\ds BAU\) \(=\) \(\ds \alpha\)
\(\ds \leadsto \ \ \) \(\ds \alpha + \map \arg {b - a}\) \(=\) \(\ds \map \arg {u - a}\)
\(\ds \leadsto \ \ \) \(\ds u - a\) \(=\) \(\ds \lambda e^{i \alpha} \paren {b - a}\) Product of Complex Numbers in Exponential Form
\(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds a + q \paren {b - a}\) where $q = \lambda e^{i \alpha}$
\(\ds \) \(=\) \(\ds \paren {1 - q} a + q b\)

$\Box$


Then we have:

\(\ds v - u\) \(=\) \(\ds b - a\) Opposite Sides and Angles of Parallelogram are Equal
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds u + b - a\)
\(\ds \) \(=\) \(\ds \paren {1 - q} a + q b + b - a\)
\(\ds \) \(=\) \(\ds -q a + \paren {1 + q} b\)

$\blacksquare$


Sources

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