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

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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

$\blacksquare$


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