Question

Let $\omega$ be the complex number $\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$. 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:

• A. $1$
• B. $5$
• C. $8$
• D. $9$

Answer 1

Answer: (A)

$1$

$\omega =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=\frac{-1}{2}+i\frac{\sqrt{3}}{2}$

$\therefore \omega$ is one of the cube roots of unity

So, the other roots are ${\omega }^2$ and $1$.

$\therefore 1+\omega +{\omega }^2=0$

$\left|\begin{array}{ccc}z+1& \omega & {\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$

$R_1\to R_1+R_2+R_3$

$\left|\begin{array}{ccc}z+1+\omega +{\omega }^2& \omega +z+{\omega }^2+1& {\omega }^2+1+z+\omega \\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$

$\left|\begin{array}{ccc}z& z& z\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$

If all the elements of a row or a column of a determinant is zero, value of determinant is zero.

$\Rightarrow z=0$