Let a, b, c be distinct complex numbers with $| a | = | b | = | c | = 1$ and $z_{1}, z_{2}$ be the roots of the equation $a z^{2} + b z + c = 0$ with $| z_{1} | = 1.$ Also, let P and Q points represent the complex number $z_{1}$ and $z_{2}$ in the complex plane with $∠ POQ=θ$ where O being the origin. Then

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$θ = 2 π / 3, P Q = \sqrt{3}$

$P Q = \sqrt{3}, b^{2} = a c$

$θ = π / 3, b^{2} = a c$

$b^{2} = a c, P Q = 2 \sqrt{3}$

answer is B, C.

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Detailed Solution

$| z_{1} z_{2} | = | \frac{c}{a} | = 1 and | z_{1} + z_{2} | = | - \frac{b}{a} | = 1$
So,
$(z_{1} + z_{2}) (\bar{z_{1}} + \bar{z_{2}}) {| z_{1} + z_{2} |}^{2} = 1 \Rightarrow 2 + \bar{z_{1}} z_{2} + z_{1} \bar{z_{2}} = 1 \Rightarrow 2 + \frac{z_{2}}{z_{1}} + \frac{z_{1}}{z_{2}} = 1 \Rightarrow \frac{{(z_{1} + z_{2})}^{2}}{z_{1} z_{2}} = 1 \Rightarrow \frac{b^{2}}{a^{2 a}} = \frac{c}{a} \Rightarrow b^{2} = a c$
Now, $z_{2} = z_{1} e^{iθ}$
$\Rightarrow | z_{1} + z_{2} | = | z_{1} | | 1 + e^{iθ} = 2 \cos \frac{θ}{2} | \cos \frac{θ}{2} + i \sin \frac{θ}{2} | | z_{1} + z_{2} | = 2 \cos \frac{θ}{2} = 1 \Rightarrow \frac{θ}{2} = \frac{π}{3} \Rightarrow θ = \frac{2 π}{3}$
Now,
$P Q = | z_{2} - z_{1} | = | z_{1} | | e^{i θ} - 1 | = | 2 \sin \frac{θ}{2} |$
We know that $θ = \frac{2 π}{3} .$ Therefore,
$P Q = | z_{2} - z_{1} | = \sqrt{3}$