Q. Let $\mathcal{z}_{1}\ and\ \mathcal{z}_{2}$ be the non-real roots of the equation $3\mathcal{z}^{2} + 3\mathcal{z +}b = 0.$ If the origin together with the points represented by $\mathcal{z}_{1}\ and\ \mathcal{z}_{2}$ form an equilateral triangle, then the value of $b$ is
A) 1
B) 2
C) 3
D) None of these
Explanation
(a)
$\mathcal{z}_{1} + \mathcal{z}_{2} = - 1\ \ and\ \mathcal{z}_{1}\mathcal{z}_{2} = \frac{b}{3}$
As $\mathcal{z}_{1},\ \mathcal{z}_{2}$ and origin form an equilateral triangle,
we have, $\mathcal{z}_{1}^{2} + \mathcal{z}_{2}^{2} + \mathcal{z}_{3}^{2} = \mathcal{z}_{1}\mathcal{z}_{2} + \mathcal{z}_{2}\mathcal{z}_{3} + \mathcal{z}_{3}\mathcal{z}_{1}$
$\Rightarrow \ \ \mathcal{z}_{1}^{2} + \mathcal{z}_{2}^{2} + 0 = \mathcal{z}_{1}\mathcal{z}_{2} + 0 + 0$
$\Rightarrow \ \ \left( \mathcal{z}_{1} + \mathcal{z}_{2} \right)^{2} = 3\mathcal{z}_{1}\mathcal{z}_{2}$
$\Rightarrow \ \ 1 = b$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.