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. $0$
• D. $4$

Answer 1

The correct option is A $1$

$\omega =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}$
Apply $C_1\to C_1+C_2+C_3$, we get
$\left|\begin{array}{ccc}z& \omega & {\omega }^2\\ z& z+{\omega }^2& 1\\ z& 1& z+\omega \end{array}\right|=0$
$\Rightarrow z\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& z+{\omega }^2& 1\\ 1& 1& z+\omega \end{array}\right|=0$
$\Rightarrow z\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 0& z+{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & z+\omega -{\omega }^2\end{array}\right|=0$
by $R_2\to R_2-R_1$
$R_3\to R_3-R_1$
$=(z+{\omega }^2-\omega )(z+\omega -{\omega }^2)-(1-\omega )(1-{\omega }^2)=0$
$\Rightarrow z^2=0$
$\Rightarrow$ only one solution.