The smallest positive integral value of n suc | vedclass

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(D) Let $z = \frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}$.Using $\sin 	heta = \cos(\frac{\pi}{2}-	he

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$8$

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(D) Let $z = \frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}$.Using $\sin 	heta = \cos(\frac{\pi}{2}-	heta)$ and $\cos 	heta = \sin(\frac{\pi}{2}-	heta)$,let $\alpha = \frac{\pi}{2}-\frac{\pi}{8} = \frac{3\pi}{8}$.Then $z = \frac{1+\cos \alpha + i \sin \alpha}{1+\cos \alpha - i \sin \alpha} = \frac{2 \cos^2 \frac{\alpha}{2} + 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2} - 2i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{\cos \frac{\alpha}{2} + i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - i \sin \frac{\alpha}{2}} = \frac{e^{i \alpha/2}}{e^{-i \alpha/2}} = e^{i \alpha}$.So,$z^n = e^{i n \alpha} = e^{i n (3\pi/8)} = \cos(\frac{3n\pi}{8}) + i \sin(\frac{3n\pi}{8})$.For $z^n = 1$,we require $\sin(\frac{3n\pi}{8}) = 0$ and $\cos(\frac{3n\pi}{8}) = 1$.This implies $\frac{3n\pi}{8} = 2k\pi$ for some integer $k$.$3n = 16k \implies n = \frac{16k}{3}$.For the smallest positive integer $n$,set $k=3$,which gives $n = 16$.

The smallest positive integral value of $n$ such that $\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{n} = 1$ is:

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