Let ω be the complex number (2 π)/(3)+i (2 π)/(3) . Then the number of distinct complex number z satisfying | array ccc z+1 & ω & ω^2 ω & z+ω^2 & 1 ω^2 & 1 & z+ω array |=0 is equal to

Correct answer: 0 Let A=[ array ccc 1 & ω & ω^2 ω & ω^2 & 1 ω^2 & 1 & ω array ] Now, A^2=[ array lll 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 array ] and Tr(A)=0,|A|=0 A^3=0 | array ccc z+1 & ω & ω^2 ω & z+ω^2 & 1 ω^2 & 1 & z+ω array |=|A+z I|=0 z^3=0 z=0 , the number of z satisfying the given equation is 1 .

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2010 · Paper 1 · Mathematics · Complex Number

Let ω be the complex number (2 π)/(3)+i (2 π)/(3) . Then the number...

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

Correct answer

Step-by-step solution

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 .

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