Let S = z ∈ ℂ : z ¯ = i z 2 + Re ( z ¯ ) . Then ∑ z ∈ S | z | 2 is equal to

Step-by-step solution

Given, z ¯ = i z 2 + Re ( z ¯ ) Now, let z = x + i y , so z ¯ = x - i y Now putting the value in, z ¯ = i z 2 + Re ( z ¯ ) we get, z ¯ = i z 2 + Re ( z ¯ ) ⇒ x - i y = i x 2 - y 2 + 2 i x y + x ⇒ x - i y = i x 2 - y 2 + x - 2 x y Now comparing real part we get, ⇒ x = - 2 x y ⇒ x ( 2 y + 1 ) = 0 ⇒ x = 0 , y = - 1 2   . . . . . 1 And imaginary part we get, - y = x 2 - y 2 + x   . . . . . . . 2 Now taking Case (I) when x = 0 in equation 2 we ge