# Complex Division/Examples/(1 + sin theta + i cos theta) (1 + sin theta - i cos theta)^-1

## 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

$\blacksquare$