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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 equal to
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Verified Answer
The correct answer is:
1
Let $A=\left[\begin{array}{ccc}1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega\end{array}\right]$
Now, $\quad A^2=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
and $\operatorname{Tr}(A)=0,|A|=0$
$\therefore \quad A^3=0$
$\Rightarrow\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=|A+z I|=0$
$\Rightarrow \quad z^3=0$
$\Rightarrow z=0$, the number of $z$ satisfying the given equation is 1 .
Now, $\quad A^2=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
and $\operatorname{Tr}(A)=0,|A|=0$
$\therefore \quad A^3=0$
$\Rightarrow\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=|A+z I|=0$
$\Rightarrow \quad z^3=0$
$\Rightarrow z=0$, the number of $z$ satisfying the given equation is 1 .
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