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

• A. 2
• B. 1
• C. 0.0
• D. 3

Answer 1

The correct option is C 0.0

$\omega =e^{i2\pi /3}$
$\left|\begin{array}{ccc}z+1& \omega & {\omega }^2\\ \omega & z+{\omega }^2& 1\\ {\omega }^2& 1& z+\omega \end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}z& \omega & {\omega }^2\\ z& z+{\omega }^2& 1\\ z& 1& z+\omega \end{array}\right|=0$
Operating $C_1\to C_1+C_2+C_3$ and using $1+\omega +{\omega }^2=0$
$\Rightarrow z\left[\right(z+{\omega }^2\left)\right(z+\omega )-1-\omega (z+\omega -1)+{\omega }^2(1-z-{\omega }^2\left)\right]=0$
$\Rightarrow z^3=0$
$\Rightarrow z=0$ is only solution.