For any integer k,let w_k = cos left( frac{kp | vedclass

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

Question

(B) Given $w_k = e^{i \frac{k\pi}{11}}$.Note that $|w_{m} - w_{n}| = |e^{i \frac{m\pi}{11}} - e^{i \frac{n\pi}{11}}| = |e^{i \frac{n\pi}{11}} (e^{i \f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $w_k = e^{i \frac{k\pi}{11}}$.Note that $|w_{m} - w_{n}| = |e^{i \frac{m\pi}{11}} - e^{i \frac{n\pi}{11}}| = |e^{i \frac{n\pi}{11}} (e^{i \frac{(m-n)\pi}{11}} - 1)| = |e^{i \frac{(m-n)\pi}{11}} - 1| = |\cos \frac{(m-n)\pi}{11} + i \sin \frac{(m-n)\pi}{11} - 1| = \sqrt{(\cos \frac{(m-n)\pi}{11} - 1)^2 + \sin^2 \frac{(m-n)\pi}{11}} = \sqrt{2 - 2 \cos \frac{(m-n)\pi}{11}} = 2 |\sin \frac{(m-n)\pi}{22}|$.For the numerator: $\sum_{k=1}^8 |w_{2k+1} - w_{2k}| = \sum_{k=1}^8 |e^{i \frac{(2k+1)\pi}{11}} - e^{i \frac{2k\pi}{11}}| = \sum_{k=1}^8 |e^{i \frac{2k\pi}{11}} (e^{i \frac{\pi}{11}} - 1)| = \sum_{k=1}^8 |e^{i \frac{\pi}{11}} - 1| = 8 |e^{i \frac{\pi}{11}} - 1|$.For the denominator: $\sum_{k=1}^4 |w_{3k-1} - w_{3k-2}| = \sum_{k=1}^4 |e^{i \frac{(3k-1)\pi}{11}} - e^{i \frac{(3k-2)\pi}{11}}| = \sum_{k=1}^4 |e^{i \frac{(3k-2)\pi}{11}} (e^{i \frac{\pi}{11}} - 1)| = \sum_{k=1}^4 |e^{i \frac{\pi}{11}} - 1| = 4 |e^{i \frac{\pi}{11}} - 1|$.Therefore,the ratio is $\frac{8 |e^{i \frac{\pi}{11}} - 1|}{4 |e^{i \frac{\pi}{11}} - 1|} = \frac{8}{4} = 2$.

For any integer $k$,let $w_k = \cos \left( \frac{k\pi}{11} \right) + i \sin \left( \frac{k\pi}{11} \right)$,where $i = \sqrt{-1}$. The value of the expression $\frac{\sum_{k=1}^8 |w_{2k+1} - w_{2k}|}{\sum_{k=1}^4 |w_{3k-1} - w_{3k-2}|}$ is

Similar Questions

If $S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}$,then $n(S)$ is:

If ${z_1}$ and ${z_2}$ are two complex numbers such that $\left| \frac{{z_1} - {z_2}}{{z_1} + {z_2}} \right| = 1$ and $i{z_1} = k{z_2}$,where $k \in R$,then the angle between ${z_1} - {z_2}$ and ${z_1} + {z_2}$ is

If $|Z_1|=|Z_2|=|Z_3|=1$ and $Z_1+Z_2+Z_3=0$,then the area of the triangle whose vertices are $Z_1, Z_2, Z_3$ is

If $z$ is a complex number satisfying $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4,$ then $|z|$ cannot be

The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module