sum_{n=1}^{20} left[ sin left( frac{2npi}{21} | vedclass

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Question

(C) We can rewrite the expression as: $\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} \right) - i \cos \left( \frac{2n\pi}{21} \right) \right] =

Vedclass · Accepted Answer

$i$

Vedclass · Answer

(C) We can rewrite the expression as: $\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} ight) - i \cos \left( \frac{2n\pi}{21} ight) ight] = -i \sum_{n=1}^{20} \left[ \cos \left( \frac{2n\pi}{21} ight) + i \sin \left( \frac{2n\pi}{21} ight) ight]$.Using Euler's formula,$e^{i	heta} = \cos 	heta + i \sin 	heta$,the expression becomes $-i \sum_{n=1}^{20} e^{i(2n\pi/21)}$.Let $\omega = e^{i(2\pi/21)}$. Then the sum is $-i \sum_{n=1}^{20} \omega^n$.This is a geometric series with $20$ terms,where the first term is $\omega$ and the common ratio is $\omega$.The sum is $\omega \frac{1-\omega^{20}}{1-\omega}$.Since $\omega^{21} = e^{i(2\pi)} = 1$,we have $\omega^{20} = \omega^{-1} = \frac{1}{\omega}$.Thus,the sum is $-i \left( \frac{\omega - \omega^{21}}{1-\omega} ight) = -i \left( \frac{\omega - 1}{1-\omega} ight) = -i (-1) = i$.

List-$I$ (Expression)	List-$II$ (Value)
$A$. $\omega^{1010} + \omega^{2000}$	$I$. $0$
$B$. $(1 + \omega - \omega^2)(1 - \omega + \omega^2)$	$II$. $1$
$C$. $(2 + \omega^2 + \omega^4)^5$	$III$. $-1$
$D$. $(3 + 5\omega + 3\omega^2)^3$	$IV$. $4$
	$V$. $8$

$\sum_{n=1}^{20} \left[ \sin \left( \frac{2n\pi}{21} \right) - i \cos \left( \frac{2n\pi}{21} \right) \right] = $

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