If z_1 and z_2 are two of the n^{text{th}} ro | vedclass

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

Question

(A) The $n^{	ext{th}}$ roots of unity are given by $z_r = e^{i \frac{2\pi r}{n}}$ for $r = 0, 1, \dots, n-1$.Let $z_1 = e^{i \frac{2\pi r_1}{n}}$ and

Vedclass · Accepted Answer

$4k$

Vedclass · Answer

(A) The $n^{	ext{th}}$ roots of unity are given by $z_r = e^{i \frac{2\pi r}{n}}$ for $r = 0, 1, \dots, n-1$.Let $z_1 = e^{i \frac{2\pi r_1}{n}}$ and $z_2 = e^{i \frac{2\pi r_2}{n}}$.The angle subtended by the line segment joining $z_1$ and $z_2$ at the origin is the difference of their arguments: $	heta = |\arg(z_1) - \arg(z_2)| = |\frac{2\pi r_1}{n} - \frac{2\pi r_2}{n}| = \frac{2\pi}{n} |r_1 - r_2|$.Given that this angle is a right angle,we have $\frac{2\pi}{n} |r_1 - r_2| = \frac{\pi}{2}$.This simplifies to $\frac{2}{n} |r_1 - r_2| = \frac{1}{2}$,which implies $n = 4 |r_1 - r_2|$.Since $|r_1 - r_2|$ is an integer,let $|r_1 - r_2| = k$,where $k$ is a positive integer.Therefore,$n$ must be of the form $4k$.

List-$I$	List-$II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$	$1.$ True
$Q.$ There exists a $k \in \{1, 2, \ldots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\|1-z_1\|\|1-z_2\| \ldots \|1-z_9\|}{10}$ equals	$3.$ $1$
$S.$ $1 - \sum_{k=1}^9 \cos \left(\frac{2k\pi}{10}\right)$ equals	$4.$ $2$

If $z_1$ and $z_2$ are two of the $n^{\text{th}}$ roots of unity such that the line segment joining them subtends a right angle at the origin,then for a positive integer $k$,$n$ takes the form

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+4=0$,then $\alpha^{12}+\beta^{12}=$

If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$,then $\alpha^{2011}$ is equal to

If $\omega$ is a complex cube root of unity,then $(x - y)(x\omega - y)(x\omega^2 - y) = $

The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module