If z_1 , z_2 and z_3 are distinct complex numbers such that ∣z_1 ∣=∣z_2 ∣=∣z_3 ∣=1 and (z_1 ^2 /(z_2 z_3 ))+(z_2 ^2 /(z_3 z_1 ))+(z_3 ^2 /(z_1 z_2 ))=−1 then the value of ∣z_1 +z_2 +z


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Question Number 139700 by EnterUsername last updated on 30/Apr/21

$If z_{1}, z_{2} and z_{3} are distinct complex numbers such$ $that ∣ z_{1} ∣=∣ z_{2} ∣=∣ z_{3} ∣= 1 and$ $\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} = - 1$ $then the value of ∣ z_{1} + z_{2} + z_{3} ∣ can be$ $(A) 1 / 2 (B) 3 (C) 3 / 2 (D) 2$
Commented by hawa last updated on 30/Apr/21

$B$
Commented by EnterUsername last updated on 30/Apr/21

$Welcome to the forum Hawa!$ $Right answer according to author is D .$ $Would you mind verifying your working ?$ $Thanks for your attention though!$
	Answered by Ar Brandon last updated on 02/May/21

	$Let z = z_{1} + z_{2} + z_{3} \Rightarrow \bar{z} = {\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3} = \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} = \frac{z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2}}{z_{1} z_{2} z_{3}}$ $\Rightarrow \bar{z} = \frac{z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2}}{b} \Rightarrow z_{2} z_{3} + z_{3} z_{1} + z_{1} z_{2} = b \bar{z}, b = z_{1} z_{2} z_{3} \land ∣ b ∣= 1$ $\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} = - 1 \Rightarrow \frac{z_{1}^{3} + z_{2}^{3} + z_{3}^{3}}{z_{1} z_{2} z_{3}} = - 1$ $\Rightarrow z_{1}^{3} + z_{2}^{3} + z_{3}^{3} - 3 z_{1} z_{2} z_{3} = - 4 z_{1} z_{2} z_{3}$ $\Rightarrow (z_{1} + z_{2} + z_{3}) (z_{1}^{2} + z_{2}^{2} + z_{3}^{2} - z_{1} z_{2} - z_{2} z_{3} - z_{3} z_{1}) = - 4 z_{1} z_{2} z_{3}$ $\Rightarrow (z_{1} + z_{2} + z_{3}) ({(z_{1} + z_{2} + z_{3})}^{2} - 3 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1})) = - 4 z_{1} z_{2} z_{3}$ $\Rightarrow z^{3} - 3 b \bar{z z} = - 4 \Rightarrow z^{3} = 3 b ∣ z ∣^{2} - 4$ $\Rightarrow∣ z^{3} ∣=∣ 3 b ∣ z ∣^{2} - 4 ∣= 3 ∣ z ∣^{2} - 4 if 3 ∣ z ∣^{2} - 4 > 0, (∣ b ∣= 1)$ $\Rightarrow∣ z ∣^{3} - 3 ∣ z ∣^{2} + 4 = 0, ∣ z ∣= 2 D$ $If 3 ∣ z ∣^{2} - 4 < 0 then ∣ z ∣= 1$
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