Let alpha and beta be the roots of x^2 + omeg | vedclass

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

Question

(A) The given quadratic equation is $x^2 + \omega x + \omega^2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) The given quadratic equation is $x^2 + \omega x + \omega^2 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:$x = \frac{-\omega \pm \sqrt{\omega^2 - 4\omega^2}}{2} = \frac{-\omega \pm \sqrt{-3\omega^2}}{2} = \frac{-\omega \pm \sqrt{3}i\omega}{2} = \omega \left( \frac{-1 \pm \sqrt{3}i}{2} \right)$.Since $\frac{-1 \pm \sqrt{3}i}{2}$ are the complex cube roots of unity (denoted as $\omega$ and $\omega^2$),the roots are $\alpha = \omega^2$ and $\beta = \omega^3 = 1$ (or vice versa).Given $z = \alpha^9 + i\beta^9$,we substitute the values:$z = (\omega^2)^9 + i(1)^9 = \omega^{18} + i$.Since $\omega^3 = 1$,$\omega^{18} = (\omega^3)^6 = 1^6 = 1$.Thus,$z = 1 + i$.The modulus $|z| = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

Let $\alpha$ and $\beta$ be the roots of $x^2 + \omega x + \omega^2 = 0$,where $\omega$ is an imaginary cube root of unity. If $z = \alpha^9 + i\beta^9$,then the value of $|z|$ is:

Similar Questions

The value of $(\frac{-1 + i\sqrt{3}}{2})^{18} + (\frac{-1 - i\sqrt{3}}{2})^{18}$ is

Let $\alpha, \beta$ denote the cube roots of unity other than $1$ and $\alpha \neq \beta$. Let $S = \sum_{n=0}^{\infty} (-1)^{n} \left(\frac{\alpha}{\beta}\right)^{n}$. Then the value of $S$ is

If $\omega \neq 1$ is a cube root of unity,then one root among the $7^{\text{th}}$ roots of $(1+\omega)$ is

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) = $

Evaluate: $\frac{(\cos \theta + i\sin \theta)^4}{(\sin \theta + i\cos \theta)^5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module