sum_{k=1}^6 left[ sin left(frac{2 pi k}{7}rig | vedclass

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Question

(C) We can rewrite the expression as: $\sum_{k=1}^6 -i \left[ \cos \left(\frac{2 \pi k}{7}\right) + i \sin \left(\frac{2 \pi k}{7}\right) \right]$ Usi

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$i$

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(C) We can rewrite the expression as: $\sum_{k=1}^6 -i \left[ \cos \left(\frac{2 \pi k}{7}ight) + i \sin \left(\frac{2 \pi k}{7}ight) ight]$ Using Euler's formula,$\cos 	heta + i \sin 	heta = e^{i 	heta}$,the expression becomes: $-i \sum_{k=1}^6 e^{i \frac{2 \pi k}{7}} = -i \left[ e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + \dots + e^{i \frac{12 \pi}{7}} ight]$ This is a geometric progression with first term $a = e^{i \frac{2 \pi}{7}}$ and common ratio $r = e^{i \frac{2 \pi}{7}}$ for $n = 6$ terms: $-i \left[ e^{i \frac{2 \pi}{7}} \frac{1 - (e^{i \frac{2 \pi}{7}})^6}{1 - e^{i \frac{2 \pi}{7}}} ight] = -i \left[ \frac{e^{i \frac{2 \pi}{7}} - e^{i \frac{14 \pi}{7}}}{1 - e^{i \frac{2 \pi}{7}}} ight]$ Since $e^{i \frac{14 \pi}{7}} = e^{i 2 \pi} = 1$: $-i \left[ \frac{e^{i \frac{2 \pi}{7}} - 1}{1 - e^{i \frac{2 \pi}{7}}} ight] = -i (-1) = i$

$\sum_{k=1}^6 \left[ \sin \left(\frac{2 \pi k}{7}\right) - i \cos \left(\frac{2 \pi k}{7}\right) \right] = $

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