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If $\omega=\cos\large\frac{\pi}{n}$$+i\sin\large\frac{\pi}{n}$,then the value of $1+\omega+\omega^2+.....+\omega^{n-1}$ is :

$\begin{array}{1 1}(A)\;1+i\\(B)\;1+i\tan(\large\frac{\pi}{n})\\(C)\;1+i\cot(\large\frac{\pi}{2n})\\(D)\;1+i\cot(\large\frac{\pi}{n})\end{array} $

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$1+\omega+\omega^2+......+\omega^{n-1}=\large\frac{1-\omega^n}{1-\omega}$
$\omega^n=\cos\large\frac{n\pi}{n}$$+i\sin \large\frac{n\pi}{n}$$=-1$
$1-\omega=2\sin^2\large\frac{\pi}{2n}-$$2i\sin\large\frac{\pi}{2n}$$\cos \large\frac{\pi}{2n}$
$\Rightarrow -2i\sin\large\frac{\pi}{2n}$$\big[\cos\large\frac{\pi}{2n}$$+i\sin \large\frac{\pi}{2n}\big]$
$\Rightarrow \large\frac{1-\omega^n}{1-\omega}=\frac{2[\cos\Large\frac{\pi}{2n}-i\sin\Large\frac{\pi}{2n}]}{-2i\sin \large\frac{\pi}{2n}}$
$\Rightarrow 1+i\cot(\large\frac{\pi}{2n})$
Hence (C) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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