sum_{r=1}^{16}left(sin frac{2 r pi}{17}+i cos | vedclass

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

Question

(D) We have,$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right) = i \sum_{r=1}^{16}\left(\cos \frac{2 r \pi}{17} - i \sin \

Vedclass · Accepted Answer

$-i$

Vedclass · Answer

(D) We have,$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right) = i \sum_{r=1}^{16}\left(\cos \frac{2 r \pi}{17} - i \sin \frac{2 r \pi}{17}\right)$ $= i \sum_{r=1}^{16} e^{-\frac{i 2 r \pi}{17}}$ Let $K = e^{-\frac{2 i \pi}{17}}$. Then the sum is $i \sum_{r=1}^{16} K^r$. This is a geometric series with $16$ terms: $i \left[ K \frac{1-K^{16}}{1-K} \right] = i \frac{K-K^{17}}{1-K}$ Since $K^{17} = e^{-2 i \pi} = \cos(2 \pi) - i \sin(2 \pi) = 1$,The expression becomes $i \frac{K-1}{1-K} = i(-1) = -i$.

$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right)=$

Similar Questions

If $|Z|=2$, $Z_1=\frac{Z}{2} e^{i \alpha}$ and $\theta$ is the $\operatorname{amp}(Z)$, then $\frac{Z_1^n-Z_1^{-n}}{Z_1^n+Z_1^{-n}}=$

If $1, \omega, \omega^2$ are the cube roots of unity,then $(1 - 2\omega + \omega^2)^6$ is equal to

Express the following expression in the form $A + iB$: $(\cos 2\theta + i\sin 2\theta )^{ - 5} (\cos 3\theta - i\sin 3\theta )^6 (\sin \theta - i\cos \theta )^3$.

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

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^9+\beta^9$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module