Question

If $z$ is a complex number satisfying $\overline{z^2}=1$, where $\overline{z}$ is the conjugate of $z$, then

• A. $Re\left(z\right)=0$
• B. $z=0$
• C. $Im\left(z\right)=0$
• D. $Re\left(z\right)=Im\left(z\right)$

Answer 1

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

Let $z=x+iy$
$\Rightarrow \overline{z}=x-iy$
Now,
$\overline{z^2}=1\Rightarrow (\overline{z}{)}^2=1(\because \overline{z^n}=\left(\overline{z}{)}^n\right)\Rightarrow (x-iy{)}^2=1\Rightarrow x^2-y^2-2ixy=1\Rightarrow x^2-y^2=1,xy=0$

When $x=0$,
$-y^2=1$
Not possible

When $y=0$,
$x^2=1\Rightarrow x=\pm 1$
Therefore,
$z=\pm 1$