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

ω = - 1 2 + 3 2 i (cube root of unity) So, 1 + ω + ω 2 = 0 and ω 3 =1 Now 1 1 1 1 ω ω 2 1 ω 2 ω = 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

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MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Hard2 minPYQ_2017

MathematicsHardsingle choice

Let $ω$ be a complex number such that $2 ω + 1 = z$ where $z = \sqrt{- 3}$ . If

$|\begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix}| = 3 k,$

Then $k$ can be equal to:

Options:

Answer:

Solution:

$ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ (cube root of unity)

So, $1 + ω + ω^{2} = 0$ and $ω^{3} =1$

Now $|\begin{matrix} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix}| = 3 (ω^{2} - ω)$

So $k = ω^{2} - ω$

$\Rightarrow k = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) - (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)$

$\Rightarrow k = - \sqrt{3} i$

$\Rightarrow k = - z$

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)

⏱ 2mℹ️ Source: PYQ_2017

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