If alpha and beta are the distinct roots of t | vedclass

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

Question

(A) The roots of the equation $x^2 - x + 1 = 0$ are $\omega$ and $\omega^2$, where $\omega$ is the complex cube root of unity.Let $\alpha = \omega$ an

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) The roots of the equation $x^2 - x + 1 = 0$ are $\omega$ and $\omega^2$, where $\omega$ is the complex cube root of unity.Let $\alpha = \omega$ and $\beta = \omega^2$.We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.Substitute the values into the expression: $\alpha^{200} + \beta^{206} + 2 = (\omega)^{200} + (\omega^2)^{206} + 2$.Simplify the powers: $\omega^{200} = \omega^{3 	imes 66 + 2} = (\omega^3)^{66} \cdot \omega^2 = 1^{66} \cdot \omega^2 = \omega^2$.$\omega^{412} = \omega^{3 	imes 137 + 1} = (\omega^3)^{137} \cdot \omega = 1^{137} \cdot \omega = \omega$.Thus, the expression becomes: $\omega^2 + \omega + 2$.Since $1 + \omega + \omega^2 = 0$, we have $\omega + \omega^2 = -1$.Therefore, $-1 + 2 = 1$.

If $\alpha$ and $\beta$ are the distinct roots of the equation $x^2 - x + 1 = 0$, then the value of $\alpha^{200} + \beta^{206} + 2$ is equal to

Similar Questions

If $\omega$ is a complex cube root of unity,then $\cos \left[\left(\omega^{1234}+\omega^{2021}\right) \pi-\frac{\pi}{4}\right]$ is equal to

If $z^2+z+1=0$,where $z$ is a complex number,then $\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3$ is equal to

One of the values of $(\sqrt{3}-i)^{\frac{1}{6}}$ is

If $\omega$ is a complex cube root of unity and $a, b, c$ are distinct real numbers,then $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} = $

For $n \in N$,if $A_n = \cos \left(\frac{\pi}{2^n}\right) + i \sin \left(\frac{\pi}{2^n}\right)$,then $(A_1 A_2 A_3 A_4)^4 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module