Question

# 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 getA¯¯¯¯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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