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If $z_{n}=\cos\large\frac{n\pi}{3}$$+i\sin\large\frac{n\pi}{3} $ then $z_{1}z_{2} \dots\;z_{6}$is

\[\begin{array}{1 1}(1)1&(2)-1\\(3)i&(4)-i\end{array}\]

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$z_n= \cos \large\frac{n \pi}{3} $$ + i \sin \large\frac{ n \pi}{3}$
$z_1.z_2-z_6 = \bigg(\cos \large\frac{\pi}{3} + $$ i \sin \large\frac{\pi}{3} \bigg) \bigg (\cos \large\frac{2 \pi}{3} $$+ i \sin \large\frac{2 \pi}{3} \bigg)$
$ \bigg(\cos \large\frac{3\pi}{3} + $$ i \sin \large\frac{3\pi}{3} \bigg) \bigg (\cos \large\frac{4 \pi}{3} $$+ i \sin \large\frac{4 \pi}{3} \bigg)$
$ \bigg(\cos \large\frac{5\pi}{3} + $$ i \sin \large\frac{5\pi}{3} \bigg) \bigg (\cos \large\frac{6 \pi}{3} $$+ i \sin \large\frac{6 \pi}{3} \bigg)$
$\qquad= \cos \bigg(\large\frac{\pi + 2 \pi +3 \pi +4 pi +5 \pi +6 \pi}{3} \bigg)+i \sin \bigg(\large\frac{\pi + 2 \pi +3 \pi +4 pi +5 \pi +6 \pi}{3} \bigg)$
$\qquad= \cos \bigg( \large\frac{21 \pi}{3} \bigg) + i \sin \bigg(\large\frac{21 \pi}{3} \bigg)$
$\qquad= \cos 7 \pi + i \sin 7 \pi$
$\qquad=-1$
Hence 2 is the correct answer.
answered May 14, 2014 by meena.p
 

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