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-o

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