Let a,b,c be distinct complex numbers with $| a | = | b | = | c | = 1 and z_{1}, z_{2}$ be the roots of the equation ${az}^{2} + bz + c = 0 with |z_{1}| = 1$ . Let P and Q represent the complex numbers z₁ and z₂ in the argand plane with $∠ POQ = θ, 0^{0} < θ < 180^{0}$ (where O being the origin) then

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$PQ = 2 \sqrt{3}; b^{2} = ac$

$θ = \frac{π}{3}; b^{2} = ac$

$θ = \frac{2 π}{3}; PQ = \sqrt{3}$

$b^{2} = ac; θ = \frac{2 π}{3}$

answer is A, B.

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Detailed Solution

$z_{1}, z_{2} are roots of the equation {az}^{2} + bz + c = 0$ , $| a | = | b | = | c | = 1$

$|z_{1} + z_{2}| = |- \frac{b}{a}| = 1, z_{1} z_{2} = \frac{c}{a} \Rightarrow |z_{1} z_{2}| = 1$

$∴ {|z_{1} + z_{2}|}^{2} = 1 \Rightarrow (z_{1} + z_{2}) (\bar{z_{1}} + \bar{z_{2}}) = 1$

$\Rightarrow \frac{{(z_{1} + z_{2})}^{2}}{z_{1} z_{2}} = 1 \Rightarrow \frac{b^{2}}{a^{2}} = \frac{c}{a}$

$\Rightarrow b^{2} = ac$

$z_{2} = z_{1} e^{iθ} \Rightarrow |z_{2} + z_{1}| = |z_{1}| |1 + e^{iθ}|$

$\Rightarrow 2 \cos (\frac{θ}{2}) = 1 \Rightarrow θ = \frac{2 π}{3}$

$P Q = |z_{2} - z_{1}| = |z_{1}| |e^{i θ} - 1| = |2 \sin \frac{θ}{2}|$

$θ = \frac{2 π}{3} \Rightarrow P Q = |z_{2} - z_{1}| = \sqrt{3}$

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