Let {z_1} and {z_2} be two roots of the equat | vedclass

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(C) Given that ${z_1}$ and ${z_2}$ are the roots of the equation ${z^2 + az + b = 0}$.By Vieta's formulas,we have ${z_1 + z_2 = -a}$ and ${z_1 z_2 = b

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${a^2 = 3b}$

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(C) Given that ${z_1}$ and ${z_2}$ are the roots of the equation ${z^2 + az + b = 0}$.By Vieta's formulas,we have ${z_1 + z_2 = -a}$ and ${z_1 z_2 = b}$.Since the origin $({z_3 = 0})$,${z_1}$,and ${z_2}$ form an equilateral triangle,the condition for the vertices ${z_1, z_2, z_3}$ to form an equilateral triangle is ${z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1}$.Substituting ${z_3 = 0}$,we get ${z_1^2 + z_2^2 = z_1 z_2}$.Adding ${2 z_1 z_2}$ to both sides,we get ${z_1^2 + z_2^2 + 2 z_1 z_2 = 3 z_1 z_2}$.This simplifies to ${(z_1 + z_2)^2 = 3 z_1 z_2}$.Substituting the values from Vieta's formulas,we get ${(-a)^2 = 3b}$,which implies ${a^2 = 3b}$.

Let ${z_1}$ and ${z_2}$ be two roots of the equation ${z^2 + az + b = 0}$,where ${z}$ is a complex number. Further,assume that the origin,${z_1}$,and ${z_2}$ form an equilateral triangle. Then:

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