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(A) The cube roots of unity are $1, \omega, \omega^2$. These roots can be represented as points on the Argand plane: $1 = (1, 0)$,$\omega = (\cos \fra

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Equilateral triangle

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(A) The cube roots of unity are $1, \omega, \omega^2$. These roots can be represented as points on the Argand plane: $1 = (1, 0)$,$\omega = (\cos \frac{2\pi}{3}, \sin \frac{2\pi}{3})$,and $\omega^2 = (\cos \frac{4\pi}{3}, \sin \frac{4\pi}{3})$. These points lie on the unit circle $|z| = 1$ and are separated by an angle of $\frac{2\pi}{3}$ radians $(120^\circ)$ from each other. Since the points are equally spaced on the circle,they form the vertices of an equilateral triangle. Alternatively,using the property that $z_1, z_2, z_3$ form an equilateral triangle if $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$,we substitute $z_1=1, z_2=\omega, z_3=\omega^2$: $1^2 + \omega^2 + (\omega^2)^2 = 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$. And $1(\omega) + \omega(\omega^2) + \omega^2(1) = \omega + \omega^3 + \omega^2 = \omega + 1 + \omega^2 = 0$. Since both sides are equal to $0$,the triangle is equilateral.

The cube roots of unity when represented on the Argand plane form the vertices of an

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