Let S = \ z ∈ ℂ: ¯ z = i( z ^ 2 + Re ( ¯ z ))\ . Then ∑ _ z ∈ S | z...

Given,

$\overset{¯}{z} = i \left({z}^{2} + Re ( \overset{¯}{z} )\right)</p><p>Now, let$z = x + iy$, so$\overset{¯}{z} = x - iy$</p><p>Now putting the value in,$\overset{¯}{z} = i \left({z}^{2} + Re ( \overset{¯}{z} )\right)$we get,</p><p>$\overset{¯}{z} = i \left({z}^{2} + Re ( \overset{¯}{z} )\right)

$⇒ x - iy = i \left({x}^{2} - {y}^{2} + 2ixy + x\right)$

$⇒ x - iy = i\left({x}^{2} - {y}^{2} + x\right) - 2xy$

Now comparing real part we get,

$⇒ x = - 2xy ⇒ x ( 2y + 1 ) = 0$

$⇒ x = 0, y = \dfrac{- 1}{2} . . . . . \left(1\right)$

And imaginary part we get,

$- y = {x}^{2} - {y}^{2} + x . . . . . . . \left(2\right)</p><p>Now taking Case (I) when$x = 0 in $equation \left(2\right) we get,</p><p>$⇒ - y = - {y}^{2}$</p><p>$⇒ {y}^{2} - y = 0 ⇒ y = 0, 1$</p><p>So,$z = 0, i$</p><p>Now taking Case (II) when$y = $\dfrac{- 1}{2}$ in equation\left(2\right) we get,

$⇒$\dfrac{1}{2} =${x}^{2} -$\dfrac{1}{4} +$x$

$⇒ {x}^{2} + x -$\dfrac{3}{4} =$0$

$⇒ 4x + 4x - 3 = 0$

$⇒ ( 2x - 1 ) ( 2x + 3 ) = 0$

$⇒ x =$\dfrac{1}{2}$,$\dfrac{- 3}{2}

So, $z =$\dfrac{1}{2} - \dfrac{1}{2}$i,$\dfrac{- 3}{2} - \dfrac{1}{2}$i$

Now finding, $∑ | z{|}^{2} = 0 + 1 +$\dfrac{1}{2} + \dfrac{5}{2} =$4$