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

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(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 = \

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$\sqrt{2}$

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(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:

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