Cube root of Unity.
I f z 2 + z + 1 = 0 W h e r e z i s a c o m p l e x n u m b e r , t h e n f i n d v a l u e o f ( z 1 + z 1 1 ) 2 + ( z 2 + z 2 1 ) 2 + ( z 3 + z 3 1 ) 2 + . . . + ( z 6 + z 6 1 ) 2
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On solving the equation z 2 + z + 1 = 0 we get z = ω , ω 2 where ω , ω 2 are cube roots of unity.
Thus we get z = ω a n d z 1 = ω 2
⟹ z + z 1 = − 1 ; z 2 + z 2 1 = − 1 ; z 4 + z 4 1 = − 1 ; z 5 + z 5 1 = − 1 and
z 3 + z 3 1 = 2 ; z 6 + z 6 1 = 2
Thus value of ( z 1 + z 1 1 ) 2 + ( z 2 + z 2 1 ) 2 + ( z 3 + z 3 1 ) 2 + . . . + ( z 6 + z 6 1 ) 2 ) = 4 + 2 ( 2 2 ) = 1 2