If z 2 + z + 1 = 0 , where z is a complex number, prove that z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 .

Given equation: z 2 + z + 1 = 0 , where z is a complex number, so clearly ω , ω 2 are the two roots which satisfies the given equation. Let z = ω To prove: z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 Now, from LHS, we have z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 4 + 1 ω 4 2 + ω 5 + 1 ω 5 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 3 · ω + 1 ω 3 · ω 2 + ω 3 · ω 2 + 1 ω 3 · ω 2 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 + ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 ∵ ω 3 = 1 , ω 6 = 1 = ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 + 8 + ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 = ω + ω 2 2 + ω 2 + ω 2 + 8 + ω + ω 2 2 + ω 2 + ω 2 = - 1 2 + - 1 2 + 8 + - 1 2 + - 1 2 ∵ 1 + ω + ω 2 = 0 = 12 = RHS Hence, proved.