Let z = 1 + a i , be a complex number, a > 0 , such that z 3 is a real number. Then, the sum 1 + z + z 2 + … . + z 11 is equal to :

Step-by-step solution

z = 1 + a i Cubing both sides, z 3 = 1 - a 2 + 2 a i + 1 - a 2 a i - 2 a 2 z 3 = 1 - 3 a 2 + ( 3 a - a 3 ) i ∵ z 3 is a real number. a 3 - a 2 = 0   ⇒     a =   3 a > 0 1 + z + z 2 … . . z 11 = z 12 - 1 z - 1 = 1 +   3 i 12 - 1 1 +   3 i - 1 = 1 +   3 i 12 - 1 3 i 1 +   3 i 12 = 2 12 1 2 + 3 2 i 12 = 2 12 cos ⁡ π 3 + i sin ⁡ π 3 12 = 2 12   cos ⁡ 4 π + i sin ⁡ 4 π = 2 12 = 2 12 - 1 3 i = - 1365 3 i