Let the complex numbers z_1, z_2 and z_3 be t | vedclass

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(C) Let $r$ be the circumradius of the equilateral triangle and $\omega$ be the cube root of unity. Let $ABC$ be the equilateral triangle with $z_1, z

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$3z_0^2$

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(C) Let $r$ be the circumradius of the equilateral triangle and $\omega$ be the cube root of unity. Let $ABC$ be the equilateral triangle with $z_1, z_2$ and $z_3$ as its vertices $A, B$ and $C$ respectively,with circumcentre $O'(z_0)$.The vectors $O'A, O'B, O'C$ are equal in magnitude and separated by an angle of $\frac{2\pi}{3}$.Then,we can write:$z_1 - z_0 = r e^{i	heta}$$z_2 - z_0 = r e^{i(	heta + \frac{2\pi}{3})} = r \omega e^{i	heta}$$z_3 - z_0 = r e^{i(	heta + \frac{4\pi}{3})} = r \omega^2 e^{i	heta}$Thus,$z_1 = z_0 + r e^{i	heta}$,$z_2 = z_0 + r \omega e^{i	heta}$,and $z_3 = z_0 + r \omega^2 e^{i	heta}$.Squaring and adding these gives:$z_1^2 + z_2^2 + z_3^2 = (z_0 + r e^{i	heta})^2 + (z_0 + r \omega e^{i	heta})^2 + (z_0 + r \omega^2 e^{i	heta})^2$$= 3z_0^2 + 2 z_0 r e^{i	heta} (1 + \omega + \omega^2) + r^2 e^{i2	heta} (1 + \omega^2 + \omega^4)$Since $1 + \omega + \omega^2 = 0$ and $1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$,we get:$z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.

Let the complex numbers $z_1, z_2$ and $z_3$ be the vertices of an equilateral triangle. Let $z_0$ be the circumcentre of the triangle,then $z_1^2 + z_2^2 + z_3^2 = $

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