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  • Please,please help me - Complex numbers and quadratic equations - JEE Main-9

 Let ω be a complex number such that

2w+1=z where z= \sqrt{-3}  if

then k is equal to :

  • Option 1)

    z

  • Option 2)

    -1

  • Option 3)

    1

  • Option 4)

    -z

 

Answers (3)

best_answer

We have, 2\omega +1=\sqrt{3}i

\Rightarrow \omega =\frac{\sqrt{3}i-1}{2}

Using, Multiplication of Complex Numbers -

(a+ib)(c+id)=(ac-bd)+i(bc+ad)

-  

\Rightarrow \omega^{2}=\frac{-2-2\sqrt{3}i}{4}\Rightarrow \frac{-1-\sqrt{3}i}{2}

\Rightarrow \omega ^{3}=-\frac{(\sqrt{3}i+1)(\sqrt{3}i-1)}{4}=1

In order to find out k, we have to find the value of:

\begin{vmatrix} 1 &1 & 1\\ 1& -\omega ^{2}-1 &\omega^{2} \\ 1& \omega^{2} & \omega^{7} \end{vmatrix} 

Here, -\omega^{2}-1=\frac{-1+\sqrt{3}i}{2}=\omega

So, \begin{vmatrix} 1 &1 & 1\\ 1& -\omega ^{2}-1 &\omega^{2} \\ 1& \omega^{2} & \omega^{7} \end{vmatrix}=\begin{vmatrix} 1 &1 & 1\\ 1& \omega &\omega^{2} \\ 1& \omega^{2} & \omega \end{vmatrix}

After expansion of determinant, we get,

\omega ^2-\omega+\omega ^2-\omega+\omega ^2-\omega=3(\omega ^2-\omega)=\omega ^2-\omega+\omega ^2-\omega+\omega ^2-\omega=3(\omega ^2-\omega)=-3\sqrt{3}i

Therefore, k=-\sqrt{3}i=-z

 

 

 

 


Option 1)

z

This option has negative sign missing. So, it is wrong.

Option 2)

-1

Not correct.

Option 3)

1

Not correct.

Option 4)

-z

This is the correct answer.

Posted by

solutionqc

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

answer is option 2 -1

Posted by

Shruti Aggarwal

View full answer

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