Sum of squares of modulus of all the complex numbers z satisfying ¯...

Given,

$z + \overset{¯}{z} = {izz}^{2} + {z}^{2}$

Consider $z = x + iy$

$2x = \left(i + 1\right)\left({x}^{2} - {y}^{2} + 2xyi\right)$

On comparing real and imaginary part we get,

$⇒ 2x = {x}^{2} - {y}^{2} - 2xy$ and ${x}^{2} - {y}^{2} + 2xy = 0$

$⇒ 2x = - 4xy$

$⇒ x = 0$ or $y = \dfrac{- 1}{2}$

Case 1: $x = 0 ⇒ 2 \times 0 = \left(i + 1\right)\left({0}^{2} - {y}^{2} + 2 \times 0 \times yi\right)y = 0$

Here $z = 0$

Case 2: $y = \dfrac{- 1}{2}$

$⇒ 4{x}^{2} - 4x - 1 = 0$

${\left(2x - 1\right)}^{2} = 2$

$2x - 1 = \pm \sqrt{2}$

$x = \dfrac{1 \pm \sqrt{2}}{2}$

Here $z = \dfrac{1 + \sqrt{2}}{2} - \dfrac{i}{2}$ or $z = \dfrac{1 - \sqrt{2}}{2} - \dfrac{i}{2}$

Sum of squares of modulus of $z$

$= 0 + \dfrac{{\left(1 + \sqrt{2}\right)}^{2} + 1}{4} + \dfrac{{\left(1 - \sqrt{2}\right)}^{2} + 1}{4} = \dfrac{8}{4} = 2$