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Properties of Conjugate of a Complex Number

The roots Z 1...

Question

The roots $Z_{1}, Z_{2}, Z_{3}$ of the equation $x^{3} + 3 p x^{2} + 3 q x + r = 0$ (p, q, r are complex numbers) correspond to
points A, B and C, then triangle ABC is equilateral, if

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Solution

The correct option is C
The triangle with vertices $A (z_{1}), B (z_{2}) a n d C (z_{3})$ is equilateral
$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2} \Rightarrow (z_{1} + z_{2} + z_{3})^{2} = 3 (z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2}) z_{1}, z_{2}, z_{3} a r e r o o t s o f x^{3} + 3 p x^{2} + 3 q x + r = 0 ∴ z_{1} + z_{2} + z_{3} = - 3 p a n d z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2} = 3 q ∴ E q . (i) \Rightarrow (- 3 p)^{2} = 3 (3 q) \Rightarrow p^{2} = q$

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