Sol. Add the given results we get

3z 1 + z 2 (1 + ω + ω 2 ) + z 3 (1 + ω 2 + ω) = A+ B + C

z 1 =

Ist multiply 2 nd & 3 rd relation by ω 2 & ω respectively and then add in I st relation we get

z 2 =

And z 3 =

|A| 2 = (z 1 + z 2 + z 3 )

= |z 1 | 2 + |z 2 | 2 + |z 3 | 2 + z 2 + z 3 + z 1 + z 3 + z 1 + z 2

|B| 2 = (z 1 + z 2 ω + z 3 ω 2 ) ( + ω 2 + ω) (  = ω 2 and = ω)

= |z 1 | 2 + |z 2 | 2 + |z 3 | 2 + z 1 ω 2 + z 3 ω + z 1 ω + z 2 ω 2 + z 2 ω + z 3 ω 2

similarly |C| 2 can be obtained

so |A| 2 + |B| 2 + |C| 2 = 3(|z 1 | 2 + |z 2 | 2 + |z 3 | 2 ) (  1 + ω + ω 2 = 0)

multiply z 1 , z 2 , z 3 we get

──────────────────────────────────────────────────────────────────────────────────────────