The cube roots of unity are the vertices of a | vedclass

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(B) The cube roots of unity are given by $1, \omega, \omega^2$,where $\omega = e^{i \frac{2\pi}{3}}$.These points lie on the unit circle $|z| = 1$ in

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Equilateral triangle

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(B) The cube roots of unity are given by $1, \omega, \omega^2$,where $\omega = e^{i \frac{2\pi}{3}}$.These points lie on the unit circle $|z| = 1$ in the complex plane.The distance between any two roots is $|1 - \omega| = |1 - (-\frac{1}{2} + i\frac{\sqrt{3}}{2})| = |\frac{3}{2} - i\frac{\sqrt{3}}{2}| = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3}$.Since the distance between all pairs of vertices is equal to $\sqrt{3}$,the triangle formed is an equilateral triangle.Thus,the correct option is $B$.

The cube roots of unity are the vertices of a/an ......... which is inscribed in a circle of unit radius,with its centre at the origin.

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