JEE Advanced · Mathematics · 3. Complex Numbers
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
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
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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