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# Given z1+z2+z3=A,z1+z2ω+z3ω2=B,z1+z2ω3+z3ω=CProve : |A|2+|B|2+|C|2=3(|z1|2+|z2|2+|z3|2)

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Solution

## We are given z1+z2+z3=Az1+z2ω+z3ω2=Bz1+z2ω2+z3ω=CWe 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+z3ω3)(¯z1+¯z2ω+¯z3ω2)=(z1+z2ω+z3ω3)(¯z1+¯z2ω2+¯z3ω)[∵¯ω=ω2and(ω)2=ω]=z1¯z1+z2¯z2ω3+z3¯z3ω.3+¯z1(z2ω+z3ω2)+¯z2(z3ω4+z1ω2)+¯z3(z1ω+z2ω2)=|z1|2+|z2|2+|z3|2+¯z1(z2ω+z3ω2)+¯z2(z3ω+z1ω2)+¯z3(z1ω+z2ω2)Similarly , C¯C=|z1|2+|z2|2+|z3|2+¯z1(z2ω2+z3ω)+¯z2(z3ω2+z1ω)+¯z3(z1ω2+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+ω2+ω)]+¯z3[z1(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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