Question

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

• A. $z$ is purely real
• B. $z$ is purely imaginary
• C. either $z$ is purely real or purely imaginary
• D. None of these

Answer 1

Answer: (C)

Let $z=x+$ iy, then its conjugate $\bar{z}=x-i y$
Given that $z^{2}=(\bar{z})^{2}$
$\Rightarrow x^{2}-y^{2}+2 i x y=x^{2}-y^{2}-2 i x y $
$\Rightarrow 4 i x y=0$
If $x \neq 0$, then $y=0$ and if $y \neq 0$, then $x=0$