If z = x + iy, xy ≠ 0 , satisfies the equation z ^ 2 + i ¯ z = 0 ,...

Given: ${z}^{2} + i z = 0$

$⇒ {z}^{2} = - i z$

Now taking $z = x + iy$ we get,

$⇒ {x}^{2} - {y}^{2} + 2ixy = - ix - y$

Now on comparing both side we get,

$⇒ {x}^{2} - {y}^{2} = - y \text{ and } 2xy = - x$

$⇒ {x}^{2} - {y}^{2} = - y \text{ and } x\left(2y + 1\right) = 0$

$⇒ {x}^{2} - {\left(\dfrac{- 1}{2}\right)}^{2} = - \left(\dfrac{- 1}{2}\right) \text{ and } y = \dfrac{- 1}{2}$

$⇒ {x}^{2} = \left(\dfrac{3}{4}\right) \text{ and } y = \dfrac{- 1}{2}$

$⇒ x = \pm \dfrac{\sqrt{3}}{2} \text{ and } y = \dfrac{- 1}{2}$

Hence, $z = \pm \dfrac{\sqrt{3}}{2} + i\left(\dfrac{- 1}{2}\right)$

$⇒ {z}^{2} = \dfrac{3}{4} - \dfrac{1}{4} \pm 2i\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)$

$⇒ {z}^{2} = \dfrac{1}{2} \pm i\left(\dfrac{\sqrt{3}}{2}\right)$

$⇒ \left|{z}^{2}\right| = {\left(\sqrt{\dfrac{1}{4} + \dfrac{3}{4}}\right)}^{2} = 1$