If z^2 + z + 1 = 0,where z is a complex numbe | vedclass

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

Question

(D) Given $z^2 + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^2$,where $\omega$ is a cube root of unity.Since $\omega^3 = 1$ and $1 + \omega +

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Given $z^2 + z + 1 = 0$,the roots are $z = \omega$ or $z = \omega^2$,where $\omega$ is a cube root of unity.Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,we have $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.Calculating each term:$1$. $\left( z + \frac{1}{z} \right)^2 = (-1)^2 = 1$$2$. $\left( z^2 + \frac{1}{z^2} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$$3$. $\left( z^3 + \frac{1}{z^3} \right)^2 = (1 + 1)^2 = 2^2 = 4$$4$. $\left( z^4 + \frac{1}{z^4} \right)^2 = (\omega + \omega^2)^2 = (-1)^2 = 1$$5$. $\left( z^5 + \frac{1}{z^5} \right)^2 = (\omega^2 + \omega)^2 = (-1)^2 = 1$$6$. $\left( z^6 + \frac{1}{z^6} \right)^2 = (1 + 1)^2 = 2^2 = 4$Sum $= 1 + 1 + 4 + 1 + 1 + 4 = 12$.

If $z^2 + z + 1 = 0$,where $z$ is a complex number,then the value of $\left( z + \frac{1}{z} \right)^2 + \left( z^2 + \frac{1}{z^2} \right)^2 + \left( z^3 + \frac{1}{z^3} \right)^2 + \dots + \left( z^6 + \frac{1}{z^6} \right)^2$ is

Similar Questions

Let $\alpha = \frac{-1 + i \sqrt{3}}{2}$. If $a = (1 + \alpha) \sum_{k=0}^{100} \alpha^{2k}$ and $b = \sum_{k=0}^{100} \alpha^{3k}$,then $a$ and $b$ are the roots of the quadratic equation:

If $1, \omega$ and $\omega^2$ are the cube roots of unity,then $(a+b+c)(a+b \omega+c \omega^2)(a+b \omega^2+c \omega) = $

If $\omega$ represents a cube root of unity and $\sum_{k=1}^n\left(k+\frac{1}{\omega}\right)\left(k+\frac{1}{\omega^2}\right)=340$,then $n=$

The value of $(8)^{1/3}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module