Question

If ω iS an imaginary cube root of unity such that (2 + ω)² = a+ bω, a. b ∈ R then value of a + b is

Answer 1

We have:

$(2+\omega)^2$ = a + bω

We have to find, the value of a + b = ?

Solution:

∴ $(2+\omega)^2$ = a + bω

⇒ 4 + 2(2)ω + $\omega^2$ = a + bω

⇒ 4 + 4ω + $\omega^2$ = a + bω

Comparing the values of a and b, we get

a = 4 and b = 4

∴ a + b = 4 + 4 = 8

Thus, the required option is "3) a + b = 8".

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Since $\omega \:$ is a cube root of unity

So

$1 + \omega + { \omega}^{2} = 0 \: \: \: and \: \: \: { \omega}^{3} = 1$

GIVEN

${(2 + \omega)}^{2} = a + b \omega$

TO DETERMINE

The value of a + b

EVALUATION

Here

${(2 + \omega)}^{2} = a + b \omega$

$\implies \: 4 + 4 \omega \: + {\omega}^{2} = a + b \omega$

$\because \: \: 1 + \omega \: + {\omega}^{2} = 0$

So

$4 + 4 \omega \: + {\omega}^{2} = a + b \omega \: \: \: \: \: \: gives \: \:$

$4 + 4 \omega \: - 1 - \omega = a + b \omega$

$\implies \: 3 + 3 \omega \: = a + b \omega$

Comparing real and imaginary parts we get

$a = 3 \: \: \: and \: \: \: b = 3$

Hence a + b = 3 + 3 = 6