Question

If $z \,\&\,\omega$ are two complex numbers, simultaneously satisfying the equation $z^3 + \omega^5 = 0 $ & $z^2\bar{\omega}^4 = 1$ then

• A. $z\, \& \omega$ both are imaginary numbers
• B. $z$ is purely real & $\omega$ is purely imaginary
• C. $z\, \&\,\omega$ both are purely real
• D. $\omega$ is purely real & $\omega$ is purely imaginary

Answer 1

Answer: (C)

Given, $z^2 - \bar{\omega}^4 = 1$
$\Rightarrow |z|^6 = \frac{1}{|\bar{\omega}|^4} = \frac{1}{|\omega|^4} (\because |\bar{\omega}| = |\omega|)$
$\therefore |z|^6 = \left(\frac{1}{|\bar{\omega}}|\right)^{12} = \left(\frac{1}{|\omega|}\right)^{12}\,...(i)$
Again , $z^3 + \omega^5 = 0$
$\Rightarrow z63 = -\omega^5$
$\Rightarrow |z|^6 = |\omega|^{10} \,....(ii)$
From (i) & (ii) we have $|\omega| = |z| = 1$
$\therefore \omega\bar{\omega} = z\bar{z} = 1$
Again $z^6 = (-\omega^5)^2 =\omega^{10}$
and $z^2(\bar{\omega})^4 = 1$
$\Rightarrow z^6(\bar{\omega})^{12} = 1$
$\Rightarrow \omega^{10}(\bar{\omega})^{12} = 1 $
$\Rightarrow (\omega\bar{\omega})^{10}(\bar{\omega})^2 = 1$
$\Rightarrow (\bar{\omega})^2 = 1$
$\Rightarrow \omega = \pm 1$ i.e., $\omega = 1, - 1$
Now if $\omega = -1$ then $z^3 - 1 = 0$
$\Rightarrow (z - 1) (z^2 + + 1) = 0$
$\Rightarrow z = 1$
If $\omega = 1$ then $z^3 + 1 = 0$
$\Rightarrow (z + 1)(z^2 - z + 1) = 0$
$\Rightarrow z = -1$
Thus $z = 1 \,\&\,\omega = -1 $ or $z = -1 , \omega = 1$
$\Rightarrow z\,\&\,\omega$ both are purely real numbers.