Equation relating Points of Parallelogram in Complex Plane/Mistake

Source Work

1960: Walter Ledermann: Complex Numbers:

Chapter $2$: Geometrical Representations

Exercise $7$

This mistake can be seen in the $1960$ edition as published by Routledge & Kegan Paul.

Mistake

The vertices parallelogram $ABVU$ are represented by the complex numbers $a, b, v, u$ respectively. The angle $UAB$ is equal to $\alpha$ and $\cmod {UA} = \lambda \cmod {AB}$. Prove that $u = \paren {1 - q} a + q b$ and $v = -q a + \paren {1 + q} b$, where $q = \lambda e^{i \alpha}$.

The configuration as described is:

However, because of the sense of $UAB$, this leads to:

$u = \paren {1 - q} a + q b$ and $v = -q a + \paren {1 + q} b$, where $q = \lambda e^{-i \alpha}$.

The question as it was intended to be stated should contain:

The angle $BAU$ is equal to $\alpha$ ...

See Equation relating Points of Parallelogram in Complex Plane for a solution to the correctly-stated exercise.

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $7$