The least positive integral value of n such t | vedclass

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(C) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.Using $\sin 	heta = \cos(\frac{\pi}

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$36$

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(C) Let $z = \frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$ and $\cos 	heta = \sin(\frac{\pi}{2} - 	heta)$,we have $\frac{\pi}{2} - \frac{2 \pi}{9} = \frac{5 \pi}{18}$.So,$z = \frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}$.Using half-angle formulas $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get:$z = \frac{2 \cos^2 \frac{5 \pi}{36} + i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}}{2 \cos^2 \frac{5 \pi}{36} - i 2 \sin \frac{5 \pi}{36} \cos \frac{5 \pi}{36}} = \frac{\cos \frac{5 \pi}{36} + i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36} - i \sin \frac{5 \pi}{36}} = \frac{e^{i 5 \pi / 36}}{e^{-i 5 \pi / 36}} = e^{i 5 \pi / 18}$.Given $z^n = 1$,we have $(e^{i 5 \pi / 18})^n = e^{i 5 n \pi / 18} = 1$.This implies $\frac{5 n \pi}{18} = 2 k \pi$ for some integer $k$.$n = \frac{36 k}{5}$.For $n$ to be the least positive integer,we set $k=5$,which gives $n = 36$.

The least positive integral value of $n$ such that $\left[\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right]^n=1$ is

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