Let z1- z2 and z3 be three complex numbers such that -z1- - -z2- - -z3- - 1 and z12z2z3- z22z3z1-z32z1z2-1-0- Then the sum of all possible values of -z1 - z2 - z3- is

Z_1^2z_2z_3 +z_2^2z_3z_1+z_3^2z_1z_2+1=0 ⇒ z^3_1+z^3_2+z^3_3+z_1z_2z_3=0 ⇒ z^3_1+z^3_2+z^3_3 3z_1z_2z_3= 4z_1z_2z_3 Since nbsp;a^3+b^3+c^3 3abc=(a+b+c)((a+b+c ...

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Standard XII

Mathematics

Geometrical Representation of Algebra of Complex Numbers

Let z1, z2 an...

Question

Let $z_{1}, z_{2}$ and $z_{3}$ be three complex numbers such that $| z_{1} | = | z_{2} | = | z_{3} | = 1$ and $\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} + 1 = 0.$ Then the sum of all possible values of $| z_{1} + z_{2} + z_{3} |$ is

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Solution

$\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} + 1 = 0 \Rightarrow z_{1}^{3} + z_{2}^{3} + z_{3}^{3} + z_{1} z_{2} z_{3} = 0 \Rightarrow z_{1}^{3} + z_{2}^{3} + z_{3}^{3} - 3 z_{1} z_{2} z_{3} = - 4 z_{1} z_{2} z_{3}$

Since $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) ((a + b + c)^{2} - 3 (a b + b c + a c)),$
$\Rightarrow (z_{1} + z_{2} + z_{3}) [{(z_{1} + z_{2} + z_{3})}_{1}^{2} - 3 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1})] = - 4 z_{1} z_{2} z_{3} \Rightarrow (z_{1} + z_{2} + z_{3}) [(z_{1} + z_{2} + z_{3})^{2} - 3 z_{1} z_{2} z_{3} (\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}})] = - 4 z_{1} z_{2} z_{3} \Rightarrow (z_{1} + z_{2} + z_{3})^{3} - 3 z_{1} z_{2} z_{3} (z_{1} + z_{2} + z_{3}) (_{1} +_{2} +_{3}) = - 4 z_{1} z_{2} z_{3} \Rightarrow | z_{1} + z_{2} + z_{3} |^{3} = | z_{1} z_{2} z_{3} | [3 | z_{1} + z_{2} + z_{3} |^{2} - 4 |]$

Let $| z_{1} + z_{2} + z_{3} | = x$
Then, $x^{3} = | 3 x^{2} - 4 |$
$\Rightarrow x = 1$ or $2$
Sum $= 1 + 2 = 3$

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Similar questions

Q. Let

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Let z1,z2 and z3 be three complex numbers such that |z1|=|z2|=|z3|=1 and z21z2z3+z22z3z1+z23z1z2+1=0. Then the sum of all possible values of |z1+z2+z3| is

Let $z_{1}, z_{2}$ and $z_{3}$ be three complex numbers such that $| z_{1} | = | z_{2} | = | z_{3} | = 1$ and $\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} + 1 = 0.$ Then the sum of all possible values of $| z_{1} + z_{2} + z_{3} |$ is