Question

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

• A. $Re\left(z\right)=Im\left(z\right)$
• B. $Re\left(z\right)Im\left(z\right)=1$
• C. $Re\left(z\right)Im\left(z\right)=0$
• D. None of these

Answer 1

The correct option is C $Re\left(z\right)Im\left(z\right)=0$

Let, $z=x+iy\Rightarrow \overline{z}=x-iy$
Given that
$z^2=(\overline{z}{)}^2\Rightarrow x^2-y^2+2ixy=x^2-y^2-2ixy\Rightarrow 4ixy=0\Rightarrow xy=0\therefore Re(z\left)Im\right(z)=0$