Q. Let $\omega_{n} = \cos{\left( \frac{2\pi}{n} \right) + i\sin{\left( \frac{2\pi}{n} \right),\ i^{2} = - 1,}}$ then $(x + y\omega_{3}\mathcal{+ z}\omega_{3}^{2})(x + y\omega_{3}^{2}\mathcal{+ z}\omega_{3})$ is equal to

A) 0
B) $x^{2} + y^{2} + \mathcal{z}^{2}$
C) $x^{2} + y^{2} + \mathcal{z}^{2} - y\mathcal{z - z}x - xy$
D) $x^{2} + y^{2} + \mathcal{z}^{2} + y\mathcal{z + z}x + xy$
Explanation

(c)
We have, $\omega_{n} = \cos{\left( \frac{2\pi}{n} \right) + i\sin\left( \frac{2\pi}{n} \right)}$
$\Rightarrow \ \ \omega_{3} = \cos{\frac{2\pi}{3} + i\sin\frac{2\pi}{3}}$
$= - \frac{1}{2} + \frac{i\sqrt{3}}{2} = \omega$
and $\omega_{3}^{2} = \left( \cos{\frac{2\pi}{3} + i\sin\frac{2\pi}{3}} \right)^{2}\$
$= \cos{\frac{4\pi}{3} + i\sin\frac{4\pi}{3}}\$
$= - \frac{1}{2} - \frac{i\sqrt{3}}{2} = \omega^{2}$
$\therefore\ \ (x + y\omega_{3}\mathcal{+ z}\omega_{3}^{2})(x + y\omega_{3}^{2}\mathcal{+ z}\omega_{3})$
$= \left( x + y\omega + \mathcal{z}\omega^{2} \right)(x + y\omega^{2}\mathcal{+ z}\omega)$
$= x^{2} + y^{2} + \mathcal{z}^{2} - xy - y\mathcal{z - z}x$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.