If the complex numbers z_1, z_2 and the origi | vedclass

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(A) Let the vertices of the triangle be $O(0), A(z_1),$ and $B(z_2)$. Since $\Delta OAB$ is an equilateral triangle,the rotation of vector $OA$ by $60

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$z_1 z_2$

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(A) Let the vertices of the triangle be $O(0), A(z_1),$ and $B(z_2)$. Since $\Delta OAB$ is an equilateral triangle,the rotation of vector $OA$ by $60^\circ$ ($\pi/3$ radians) about the origin $O$ must coincide with the vector $OB$ (or vice versa).Thus,we have $z_2 = z_1 e^{\pm i\pi/3}$.This implies $\frac{z_2}{z_1} = \cos(\pi/3) \pm i \sin(\pi/3) = \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$.Rearranging gives $2z_2 = z_1(1 \pm i\sqrt{3})$,or $2z_2 - z_1 = \pm i\sqrt{3} z_1$.Squaring both sides: $(2z_2 - z_1)^2 = -3z_1^2$.$4z_2^2 - 4z_1 z_2   z_1^2 = -3z_1^2$.$4z_1^2   4z_2^2 = 4z_1 z_2$.Dividing by $4$,we get $z_1^2   z_2^2 = z_1 z_2$.

If the complex numbers $z_1, z_2$ and the origin form an equilateral triangle,then $z_1^2 + z_2^2 = $

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