If ω (≠ 1) is a cube root of unity and (1 + ω)^7 = A + Bω, then A and B are respectively

If ω (≠ 1) is a cube root of unity and (1 + ω)^7 = A + Bω,… — JEE Main 1995 Mathematics PYQ

1995

\omega

(

\neq 1

) is a cube root of unity and

(1 + \omega)^7 = A + B\omega

, then

A

and

B

are respectively

A.
0, 1
B.
1, 1
C.
1, 0
D.
$-1$ , 1

Show answer and solution

Correct option: B

Solution

Since ω is a primitive cube root of unity, 1 + ω = -ω². So (1 + ω)⁷ = (-ω²)⁷ = -ω¹⁴. Now ω³ = 1, so ω¹⁴ = ω^(14 mod 3) = ω², giving -ω² = -(ω²). Since 1 + ω + ω² = 0, we have -ω² = 1 + ω. Therefore A + Bω = 1 + ω, giving A = 1, B = 1.

[1995]

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JEE Main 1995 Mathematics question on Complex Numbers and Quadratic Equations