Question

The number of complex numbers which are conjugate of their own cube, is ___

Answer 1

Let $z$ be such a complex number.
Then according to the question, we have
$ z^{3}=\bar{z} \,\,\,....(1)$
$\Rightarrow |z|=0 $
or $|z|=1$
$\Rightarrow z=0$
or $ z=\cos \theta+i \sin \theta$
Putting $z=\cos \theta+i \sin \theta$ in (1),
we get $\cos 3 \theta=\cos \theta$
and $\sin 3 \theta=-\sin \theta$
$\Rightarrow \sin \theta=0$
or $\sin \theta=1$
or $\sin \theta=-1$
$\Rightarrow \theta=0$
or $\pi$, or $\frac{\pi}{2}$
or $-\frac{\pi}{2}$
$\Rightarrow z=1,-1, i,-i$
or $z=0$