{left( {frac{{1 + sin theta + icos theta }}{{ | vedclass

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Question

(A) Let $z = \frac{1 + \sin 	heta + i\cos 	heta}{1 + \sin 	heta - i\cos 	heta}$.Using the identities $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$

Vedclass · Accepted Answer

$\cos \left( {\frac{{n\pi }}{2} - n	heta } ight) + i\sin \left( {\frac{{n\pi }}{2} - n	heta } ight)$

Vedclass · Answer

(A) Let $z = \frac{1 + \sin 	heta + i\cos 	heta}{1 + \sin 	heta - i\cos 	heta}$.Using the identities $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$ and $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$, let $\alpha = \frac{\pi}{2} - 	heta$.Then $z = \frac{1 + \cos \alpha + i\sin \alpha}{1 + \cos \alpha - i\sin \alpha}$.Using half-angle formulas $1 + \cos \alpha = 2\cos^2(\frac{\alpha}{2})$ and $\sin \alpha = 2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})$:$z = \frac{2\cos^2(\frac{\alpha}{2}) + i(2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2}))}{2\cos^2(\frac{\alpha}{2}) - i(2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2}))} = \frac{\cos(\frac{\alpha}{2}) + i\sin(\frac{\alpha}{2})}{\cos(\frac{\alpha}{2}) - i\sin(\frac{\alpha}{2})}$.Using Euler's formula $e^{ix} = \cos x + i\sin x$, this is $\frac{e^{i\alpha/2}}{e^{-i\alpha/2}} = e^{i\alpha}$.Thus, $z^n = (e^{i\alpha})^n = e^{in\alpha} = \cos(n\alpha) + i\sin(n\alpha)$.Substituting $\alpha = \frac{\pi}{2} - 	heta$, we get $\cos(n(\frac{\pi}{2} - 	heta)) + i\sin(n(\frac{\pi}{2} - 	heta)) = \cos(\frac{n\pi}{2} - n	heta) + i\sin(\frac{n\pi}{2} - n	heta)$.

${\left( {\frac{{1 + \sin \theta + i\cos \theta }}{{1 + \sin \theta - i\cos \theta }}} \right)^n} = $

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