If ω( 1) is a cube root of unity and (1+ω)⁷=A+Bω, then (A,B) equals (a) (1,1) (b) (1,0) (c) (-1,1) (d) (0,1)

Correct answer: (a) (1,1) 1+ω=-ω², so (1+ω)⁷=-ω¹⁴=-ω²=1+ω. Hence A+Bω=1+ω→(A,B)=(1,1). JEE Main 2011 · Complex Numbers — verified solution by the Vidaara Team.

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[JEE Main 2011] if omega 1 is a cube root of unity and 1 omega 7 a

VAVidaara Admin Asked 2d ago 0 views 1 answer

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If $\omega(\ne1)$ is a cube root of unity and $(1+\omega)^7=A+B\omega$, then $(A,B)$ equals

(a) $(1,1)$
(b) $(1,0)$
(c) $(-1,1)$
(d) $(0,1)$

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Correct answer: (a) $(1,1)$

$1+\omega=-\omega^2$, so $(1+\omega)^7=-\omega^{14}=-\omega^2=1+\omega$. Hence $A+B\omega=1+\omega\Rightarrow(A,B)=(1,1)$.

JEE Main 2011 · Complex Numbers — verified solution by the Vidaara Team.

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