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 $\begin{vmatrix}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{vmatrix}=0$ is

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (A)

Denote the given determinant by $\Delta$.
Using $C_1 \rightarrow C_1+C_2+C_3$ and $1+\omega+\omega^2=0$, we get
$\Delta=z\begin{vmatrix}1 & \omega & \omega^2 \\1 & z+\omega^2 & 1 \\1 & 1 & z+\omega\end{vmatrix}$
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get
$\Delta =z\begin{vmatrix}1 & \omega & \omega^2 \\0 & z+\omega^2-\omega & 1-\omega^2 \\0 & 1-\omega & z+\omega-\omega^2\end{vmatrix}$
$=z\left[\left(z+\omega^2-\omega\right)\left(z+\omega-\omega^2\right)-(1-\omega)\left(1-\omega^2\right)\right]$
$ =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.