Question

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

The correct option is C

Either z is purely real or purely imaginary

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