Question

Let $\omega$ be the complex number $\cos \frac{2\pi }{3}$ $+i\sin \frac{2\pi }{3}$. Then the number of distinct complex number $z$ 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

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (A)

Denote the given determinant by $\Delta$.
Using $C_1\to C_1+C_2+C_3$ and $1+\omega +{\omega }^2=0$, we get
$\Delta =z\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& z+{\omega }^2& 1\\ 1& 1& z+\omega \end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we get
$\Delta =z\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 0& z+{\omega }^2-\omega & 1-{\omega }^2\\ 0& 1-\omega & z+\omega -{\omega }^2\end{array}\right|$
$=z\left[\left(z+{\omega }^2-\omega \right)\left(z+\omega -{\omega }^2\right)-(1-\omega )\left(1-{\omega }^2\right)\right]$
$=z\left[z^2-{\left({\omega }^2-\omega \right)}^2-\left(1-\omega -{\omega }^2+1\right)\right]$
$=z\left[z^2-\left({\omega }^4+{\omega }^2-2{\omega }^3\right)-3\right]=z\left(z^2\right)=z^3$
Thus, $z^3=0\Rightarrow z=0$
Thus, there is just one value of $z$, satisfying the given equation.