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(1+ω)⁷=l+mω Find l+m | Cube Roots of Unity JEE

JEE Maths question with a full step-by-step solution.

Question

\omega \neq 1

is a cube root of unity and

(1 + \omega)^7 = l + m\omega

, then the value of

l + m

0

1

2

correct

-1

Solution

Step 1: Use the defining identity of cube roots of unity,

1 + \omega + \omega^2 = 0

, which gives:

1 + \omega = -\omega^2

Step 2: Raise to the seventh power:

(1 + \omega)^7 = (-\omega^2)^7 = (-1)^7 (\omega^2)^7 = -\omega^{14}

Step 3: Reduce

\omega^{14}

using

\omega^3 = 1

. Since

14 = 3\cdot 4 + 2

\omega^{14} = (\omega^3)^4\cdot\omega^2 = 1^4\cdot\omega^2 = \omega^2

Hence

(1 + \omega)^7 = -\omega^2

. Step 4: Express

-\omega^2

in the form

l + m\omega

. From Step 1,

-\omega^2 = 1 + \omega

, so:

(1 + \omega)^7 = 1 + \omega = l + m\omega \implies l = 1, \quad m = 1

Step 5: Therefore

l + m = 1 + 1 = 2

Correct answer: (3)

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