Let ω be a complex number such that 2ω + 1 = z where z = - 3 . If | cases 1 & 1 & 1 1 & - ω ^ 2 - 1 & ω ^ 2 1 & ω ^ 2 & ω ^ 7 cases | = 3k, Then k can be equal to:

Correct answer: A. – z ω = - (1)/(2) + 3 2 i (cube root of unity) So, 1 + ω + ω ^ 2 = 0 and ω ^ 3 =1 Now | cases 1 & 1 & 1 1 & ω & ω ^ 2 1 & ω ^ 2 & ω cases | = 3( ω ^ 2 - ω) So k = ω ^ 2 - ω ⇒ k = (- (1)/(2) - 3 2 i) - (- (1)/(2) + 3 2 i) ⇒ k = - 3 i ⇒ k = - z

JEE Main 20172 April

Exam: JEE Main
Year: 2017
Date: 2 April
Subject: Mathematics
Chapter: Complex Number

JEE Main 2017 · 2 April · Mathematics · Complex Number

Let ω be a complex number such that 2ω + 1 = z where z = - 3 . If |...

Let

\omega be a complex number such that

2\omega + 1 = z

where

z = \sqrt{- 3}

. If<br /><br />\left|

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

\right| = 3k,

Then

k

can be equal to:

Options

A $– z$
B $\dfrac{1}{z}$
C $- 1$
D $1$

Correct answer

A. $– z$

Step-by-step solution

\omega = - \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i $(cube root of unity)<br /><br />So,$ 1 + \omega + {\omega}^{2} = 0 $and$ {\omega}^{3} =1 $<br /><br />Now \left|$ \begin{cases}1 & 1 & 1 \\ 1 & \omega & {\omega}^{2} \\ 1 & {\omega}^{2} & \omega\end{cases} $\right| = 3\left({\omega}^{2} - \omega\right)$

So $k = {\omega}^{2} - \omega$

$⇒ k = \left(- \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) - \left(- \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)$

$⇒ k = - \sqrt{3}i$

$⇒ k = - z$

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