Solve for $z w h e r e ¯ ¯ ¯ z = i z^{2}$

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Solution

$¯ z = i z^{2}, z = x + i y$

$x - i y = i (x^{2} - y^{2} + 2 i x y)$

$= i (x^{2} - y^{2}) - 2 x y$

$\Rightarrow x = - 2 x y$ & $- y = x^{2} - y^{2}$

$\Rightarrow x + 2 x y = 0$ & $x^{2} - y^{2} + y = 0$

$\Rightarrow x (1 + 2 y) = 0$ & $x^{2} - y^{2} + y = 0$

$y = - \frac{1}{2} a n d x = 0$

$y = - 1 / 2$

$x^{2} - y^{2} + y = 0$

$\Rightarrow x^{2} - 1 / 4 - 1 / 2 = 0$

$\Rightarrow x^{2} = 3 / 4$

$x = \pm \sqrt{3 / 2}$

$\Rightarrow z = \frac{\pm \sqrt{3}}{2} - \frac{i}{2}$

$x = 0$

$x^{2} - y^{2} + y = 0$

$\Rightarrow y^{2} - y = 0$

$\Rightarrow y = 0, 1$

$\Rightarrow z = 0$

$z = i$

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