If z = frac{-1-i sqrt{3}}{2},then sum_{k=1}^{ | vedclass

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

Question

(C) Given $z = \frac{-1-i \sqrt{3}}{2} = \omega$,where $\omega$ is the complex cube root of unity. We know that $\omega^3 = 1$ and $1 + \omega + \omeg

Vedclass · Accepted Answer

$4044$

Vedclass · Answer

(C) Given $z = \frac{-1-i \sqrt{3}}{2} = \omega$,where $\omega$ is the complex cube root of unity. We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Also,$\frac{1}{\omega} = \omega^2$. The expression is $\sum_{k=1}^{2022} (\omega^k + \omega^{2k})^2 = \sum_{k=1}^{2022} (\omega^{2k} + \omega^{4k} + 2\omega^{3k})$. Since $\omega^{3k} = 1$,this becomes $\sum_{k=1}^{2022} (\omega^{2k} + \omega^k + 2)$. We can split the sum: $\sum_{k=1}^{2022} \omega^{2k} + \sum_{k=1}^{2022} \omega^k + \sum_{k=1}^{2022} 2$. Since $2022$ is a multiple of $3$,the sum of powers of $\omega$ over $3$ terms is $0$. Thus,$\sum_{k=1}^{2022} \omega^k = 0$ and $\sum_{k=1}^{2022} \omega^{2k} = 0$. The total sum is $0 + 0 + 2 	imes 2022 = 4044$.

List-$I$ (Expression)	List-$II$ (Value)
$A$. $\omega^{1010} + \omega^{2000}$	$I$. $0$
$B$. $(1 + \omega - \omega^2)(1 - \omega + \omega^2)$	$II$. $1$
$C$. $(2 + \omega^2 + \omega^4)^5$	$III$. $-1$
$D$. $(3 + 5\omega + 3\omega^2)^3$	$IV$. $4$
	$V$. $8$

If $z = \frac{-1-i \sqrt{3}}{2}$,then $\sum_{k=1}^{2022} \left(z^k + \frac{1}{z^k}\right)^2 = $

Similar Questions

If $\omega$ is an imaginary cube root of unity,then the value of $(1-\omega+\omega^{2}) \cdot(1-\omega^{2}+\omega^{4}) \cdot(1-\omega^{4}+\omega^{8}) \cdot \ldots$ ($2n$ factors) is

The value of $\left[ \frac{1 - \cos \frac{\pi}{10} + i\sin \frac{\pi}{10}}{1 - \cos \frac{\pi}{10} - i\sin \frac{\pi}{10}} \right]^{10} = $

If $\alpha$ is a complex number satisfying the equation $\alpha^{2}+\alpha+1=0$,then $\alpha^{31}$ is equal to

If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$,then $\left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module