Complex Division/Examples/(1 + sin theta + i cos theta) (1 + sin theta - i cos theta)^-1
Jump to navigation
Jump to search
Theorem
- $\dfrac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta} = \sin \theta + i \cos \theta$
Proof
\(\ds \frac {1 + \sin \theta + i \cos \theta} {1 + \sin \theta - i \cos \theta}\) | \(=\) | \(\ds \frac {\paren {1 + \sin \theta + i \cos \theta}^2} {\paren {1 + \sin \theta - i \cos \theta} \paren {1 + \sin \theta + i \cos \theta} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {1 + \sin \theta}^2 + 2 i \cos \theta \paren {1 + \sin \theta} - \cos^2 \theta} {\paren {1 + \sin \theta}^2 + \cos^2 \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 + 2 \sin \theta + \sin^2 \theta + 2 i \cos \theta \paren {1 + \sin \theta} - \cos^2 \theta} {1 + 2 \sin \theta + \sin^2 \theta + \cos^2 \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 + 2 \sin \theta + 2 \sin^2 \theta + 2 i \cos \theta \paren {1 + \sin \theta} - \paren {\cos^2 \theta + \sin^2 \theta} } {1 + 2 \sin \theta + \sin^2 \theta + \cos^2 \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \sin \theta \paren {1 + \sin \theta} + 2 i \cos \theta \paren {1 + \sin \theta} } {2 + 2 \sin \theta}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta + i \cos \theta\) | simplification |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Exercise $11$