Let z_{1} be a fixed point on the circle of r | vedclass

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(B) Given that $z_{1}, z_{2}, z_{3}$ are vertices of an equilateral triangle inscribed in the unit circle $|z| = 1$. Since the triangle is equilateral

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$z_{1}^{3}$

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(B) Given that $z_{1}, z_{2}, z_{3}$ are vertices of an equilateral triangle inscribed in the unit circle $|z| = 1$. Since the triangle is equilateral and inscribed in the circle centered at the origin, the vertices $z_{2}$ and $z_{3}$ can be obtained by rotating $z_{1}$ by $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ radians respectively. Thus, $z_{2} = z_{1} e^{i(2\pi/3)} = z_{1}\omega$ and $z_{3} = z_{1} e^{i(4\pi/3)} = z_{1}\omega^{2}$, where $\omega$ is the cube root of unity. Therefore, the product $z_{1} z_{2} z_{3} = z_{1} 	imes (z_{1}\omega) 	imes (z_{1}\omega^{2}) = z_{1}^{3} \omega^{3}$. Since $\omega^{3} = 1$, we have $z_{1} z_{2} z_{3} = z_{1}^{3}$.

Let $z_{1}$ be a fixed point on the circle of radius $1$ centered at the origin in the Argand plane and $z_{1} \neq \pm 1$. Consider an equilateral triangle inscribed in the circle with $z_{1}, z_{2}, z_{3}$ as the vertices. Then, $z_{1} z_{2} z_{3}$ is equal to

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