If z = sin theta - i cos theta, then for any | vedclass

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(A) Given,$z = \sin 	heta - i \cos 	heta$ $z = \cos \left(	heta - \frac{\pi}{2}ight) + i \sin \left(	heta - \frac{\pi}{2}ight) = e^{i(	heta -

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$z^{n} + \frac{1}{z^{n}} = 2 \cos \left(\frac{n \pi}{2} - n 	hetaight)$

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(A) Given,$z = \sin 	heta - i \cos 	heta$ $z = \cos \left(	heta - \frac{\pi}{2}ight) + i \sin \left(	heta - \frac{\pi}{2}ight) = e^{i(	heta - \frac{\pi}{2})}$ Using De Moivre's Theorem,$z^{n} = e^{i(n	heta - \frac{n\pi}{2})} = \cos \left(n 	heta - \frac{n \pi}{2}ight) + i \sin \left(n 	heta - \frac{n \pi}{2}ight)$ Then,$\frac{1}{z^{n}} = z^{-n} = e^{-i(n	heta - \frac{n\pi}{2})} = \cos \left(n 	heta - \frac{n \pi}{2}ight) - i \sin \left(n 	heta - \frac{n \pi}{2}ight)$ Adding these,$z^{n} + \frac{1}{z^{n}} = 2 \cos \left(n 	heta - \frac{n \pi}{2}ight) = 2 \cos \left(\frac{n \pi}{2} - n 	hetaight)$ Thus,option $A$ is correct.

If $z = \sin \theta - i \cos \theta,$ then for any integer $n$

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