# Equation relating Points of Parallelogram in Complex Plane/Mistake

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## 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 $7$