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. 2
• C. -1
• D. -2

Answer 1

The correct option is A 1

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

$C_1\to C_1+C_2+C_3$
$z\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& z+{\omega }^2& 1\\ 1& 1& z+\omega \end{array}\right|=0$

$\Rightarrow z\left[\right(z+{\omega }^2\left)\right(z+\omega )-1-\omega (z+\omega -1)+{\omega }^2(1-z-{\omega }^2\left)\right]=0$

$\Rightarrow z^3=0$

$\Rightarrow z=0$ is only solution