Let z = a + i b , b ≠ 0 be complex numbers satisfying z 2 = z ¯ · 2 1 - z . Then the least value of n ∈ N , such that z n = z + 1 n , is equal to _____ .

Step-by-step solution

Given, z 2 = z ¯ · 2 1 - z           ⋯ 1 On taking modulus both side we get, ⇒ z 2 = z · 2 1 - z ⇒ z = 2 1 - z ,    ∵   b ≠ 0 ⇒ z ≠ 0 So comparing both side we get z = 1         ⋯ 2 Now putting z = a + i b then a 2 + b 2 = 1         ⋯ 3 Now again from equation 1 , equation 2 , equation 3 we get: a 2 - b 2 + i 2 a b = a - i b 2 0 = a - i b Now on comaparing imagenary an