sum_{k=1}^6 left[ sin frac{2 k pi}{7} - i cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have,$S = \sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ Factor out $-i$: $S = \sum_{k=1}^6 (-i) \left( \cos \f

Vedclass · Accepted Answer

$i$

Vedclass · Answer

(D) We have,$S = \sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ Factor out $-i$: $S = \sum_{k=1}^6 (-i) \left( \cos \frac{2 k \pi}{7} + i \sin \frac{2 k \pi}{7} \right)$ Let $\omega = e^{i \frac{2 \pi}{7}} = \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7}$. Then the expression becomes: $S = -i \sum_{k=1}^6 \omega^k$ This is a geometric progression sum: $S = -i \left( \omega + \omega^2 + \dots + \omega^6 \right)$ Since $\omega$ is a $7^{th}$ root of unity,$1 + \omega + \omega^2 + \dots + \omega^6 = 0$. Therefore,$\omega + \omega^2 + \dots + \omega^6 = -1$. Substituting this into the expression for $S$: $S = -i (-1) = i$.

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

Similar Questions

$n$ is a positive integer and not a multiple of $3$. If $\omega$ is a non-real cube root of unity,then $\omega^n + \omega^{2n}$ is equal to

The product of all the values of $(\sqrt{3}-i)^{\frac{3}{7}}$ is

If $z^{2} + z + 1 = 0$,$z \in \mathbb{C}$,then $\left| \sum_{n=1}^{15} \left( z^{n} + (-1)^{n} \frac{1}{z^{n}} \right)^{2} \right|$ is equal to

If $\omega$ is a complex cube root of unity,then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}=$

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module