Let z ¯ denote the complex conjugate of a complex number z and let i = - 1 . In the set of complex numbers, the number of distinct roots of the equation z ¯ - z 2 = i z ¯ + z 2 is _______.

Step-by-step solution

Given z ¯ 1 - i = z 2 1 + i So z ¯ 1 - i = z 2 1 + i ⇒ z = z 2 ⇒ z = 0 or z = 1 Also, z ¯ 1 - i = z 2 1 + i ⇒ z ¯ 1 - i 1 + i = z 2 ⇒ - i z ¯ = z 2 Let z = x + i y ⇒ x 2 - y 2 - 2 x y i = y - i x i.e. x 2 - y 2 = y and - 2 x y = - x From - 2 x y = - x , we get x = 0   or   y = - 1 2 i.e. the points are 0 , 0 , 0 , 1 , 3 2 , - 1 2 and - 3 2 , - 1 2