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If z1+z2+z3=A,z1+z2ω+z3ω2=B and z1+z2ω2+z3ω=C, where 1,ω,ω2 are the three cube root of unity, then |A|2+|B|2+|C|2=

A
3(|z1|2+|z2|2+|z3|2)
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B
2(|z1|2+|z2|2+|z3|2)
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C
(|z1|2+|z2|2+|z3|2)
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D
none of these
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Solution

The correct option is A 3(|z1|2+|z2|2+|z3|2)
We have, |A|2+|B|2+|C|2=A¯¯¯¯A+B¯¯¯¯B+C¯¯¯¯C ....(1)
But A¯¯¯¯A=(z1+z2+z3)(¯¯¯¯¯z1+¯¯¯¯¯z2+¯¯¯¯¯z3)

=z1¯¯¯¯¯z1+z2¯¯¯¯¯z2+z3¯¯¯¯¯z3+¯¯¯¯¯z1(z2+z3)+¯¯¯¯¯z2(z3+z1)+¯¯¯¯¯z3(z1+z2)

=|z1|2+|z2|2+|z3|2+¯¯¯¯¯z1(z2+z3)+¯¯¯¯¯z2(z3+z1)+¯¯¯¯¯z3(z1+z2)

B¯¯¯¯B=(z1+z2ω+z3ω2)(¯¯¯¯¯z1+¯¯¯¯¯¯¯¯z2ω+¯¯¯¯¯¯¯¯¯¯z3ω2)

=(z1+z2ω+z3ω2)(¯¯¯¯¯z1+¯¯¯¯¯z2ω2+¯¯¯¯¯¯¯¯¯¯z3ω2)

[¯¯¯ω=ω2 and ¯¯¯¯¯¯¯¯¯¯(ω2)=ω]

=z1¯¯¯¯¯z1+z2¯¯¯¯¯z2ω3+z3¯¯¯¯¯z3ω3+¯¯¯¯¯z1(z2ω+z3ω2)+¯¯¯¯¯z2(z3ω4+z1ω2)+¯¯¯¯¯z3(z2ω+z2ω2)

=|z1|2+|z2|2+|z3|2+¯¯¯¯¯z1(z2ω+z3ω2)+¯¯¯¯¯z2(z1ω+z1ω2)+¯¯¯¯¯z3(z1ω+z3ω2) ...(2)

similarly, C¯¯¯¯C=|z1|2+|z2|2+|z3|2+¯¯¯¯¯z1(z2ω2+z3ω)

+¯¯¯¯¯z2(z3ω+z1ω2)+¯¯¯¯¯z3(z2ω2+z2ω) ...(3)

adding (1),(2) and (3), we get
A¯¯¯¯A+B¯¯¯¯B+C¯¯¯¯C=3[|z1|2+|z2|2+|z3|2]

+¯¯¯¯¯z1[z2(1+ω+ω2)+z3(1+ω2+ω)]
+¯¯¯¯¯z2[z3(1+ω+ω2)+z1(1+ω+ω2)]
+¯¯¯¯¯z3[z2(1+ω+ω2)+z2(1+ω2+ω)]

=3[|z1|2+|z2|2+|z3|2][1+ω+ω2=0]
From (1) and (2), we conclude
|A|2+|B|2+|C|2=3[|z1|2+|z2|2+|z3|2].

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