If z = 1 2 3 - i and the least positive integral value of n such that z 101 + i 109 106 = z n is k , then the value of 2 5 k is equal to

Step-by-step solution

z = - 1 2 i 1 + i 3 = i ω 2 z 101 = i ω z 101 + i 109 106 = i ω + i 106 = i ( − ω 2 ) 106 = − ω 2 as given that z 101 + i 109 106 = z n ∴ - ω 2 = i ω 2 n = i n ω 2 n ω 2 n - 2 i n = - 1 this is possible only when n = 4 r + 2 and 2 n − 2 is a multiple of 3 i.e., 2 4 r + 2 - 2 is a multiple of 3 i.e., 8 r + 2 is a multiple of 3 ⇒ r = 2 ∴ n = 10 ∴ 2 5 k = 4