Sum of squares of modulus of all the complex numbers z satisfying z ¯ = i z 2 + z 2 - z is equal to

Step-by-step solution

Given, z + z ¯ = izz 2 + z 2 Consider z = x + i y 2 x = i + 1 x 2 - y 2 + 2 x y i On comparing real and imaginary part we get, ⇒ 2 x = x 2 - y 2 - 2 x y and x 2 - y 2 + 2 x y = 0 ⇒ 2 x = - 4 x y ⇒   x = 0 or y = - 1 2 Case 1: x = 0 ⇒ 2 × 0 = i + 1 0 2 - y 2 + 2 × 0 × y i y = 0 Here z = 0 Case 2: y = - 1 2 ⇒ 4 x 2 - 4 x - 1 = 0 2 x - 1 2 = 2 2 x - 1 = ± 2 x = 1 ± 2 2 Here z = 1 + 2 2 - i 2 or z = 1 - 2 2 - i 2 Sum of squares of modulus of z = 0 + 1 + 2