Question

Let $\omega$ be the complex number $cos\frac{2\pi }{3}+isin\frac{2\pi }{3}$. Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc}z+1& \omega & {\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$ is equal to

Answer 1

Given that $\omega =cos\frac{2\pi }{3}+isin\frac{2\pi }{3}$
Therefore, $1+\omega +{\omega }^2=0$ & ${\omega }^3=1$
$\left|\begin{array}{ccc}z+1& \omega & {\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$
$R_1⟶R_1+R_2+R_3$
$\left|\begin{array}{ccc}z+1+\omega +{\omega }^2& z+1+\omega +{\omega }^2& z+1+\omega +{\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$
$\Rightarrow z\left|\begin{array}{ccc}1& 1& 1\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$
$\Rightarrow z[(z+w^2)(z+w)-1-\omega (z+\omega )+{\omega }^2-{\omega }^2(z+{\omega }^2)]=0$
$\Rightarrow z[z^2+z(\omega +{\omega }^2)-z(\omega +{\omega }^2)]=0$
$\Rightarrow z^3=0$
Ans: 1