Let ω be a complex number such that 2ω+1=z where z=√(-3) . If 1 1 1 1 -ω^2-1 ω^2 1 ω^2 ω^7 =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 1 1 1 1 ω ω^2 1 ω^2 ω =3ω^2-ω So k=ω^2-ω ⇒k=-1/2-√32i--1/2+√32i ⇒k=-√3i ⇒k=-z

Q35 / 90 · JEE Main 2017 (02 Apr) · Mathematics

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:

Correct answer
A
$- z$
B
$\frac{1}{z}$
C
$- 1$
D
$1$

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 $k = ω^{2} - ω$

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

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

$\Rightarrow k = - z$

Hence, the correct answer is option: 'A'.

Let ω be a complex number such that 2ω+1=z where z=-3 . If1111-ω2-1ω21ω2ω7=3k,Then k can be equal to:

Solution

Relevant Questions

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: