If $z$ is a complex number such that $z^{2} = (¯ ¯ ¯ z)^{2}$ , then

R e (z) = I m (z)

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R e (z) I m (z) = 1

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R e (z) I m (z) = 0

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None of these

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Solution

The correct option is C $R e (z) I m (z) = 0$
Let, $z = x + i y \Rightarrow ¯ ¯ ¯ z = x - i y$
Given that
$z^{2} = (¯ ¯ ¯ z)^{2} \Rightarrow x^{2} - y^{2} + 2 i x y = x^{2} - y^{2} - 2 i x y \Rightarrow 4 i x y = 0 \Rightarrow x y = 0 ∴ R e (z) I m (z) = 0$

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