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.