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Complex Numbers question

2011 · Shift 0 · Q48
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Complex Numbers question

2011 · Shift 0 · Q48

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If ω(e1)\omega ( e 1)ω(e1) is a cube root of unity, and (1+ω)7=A+Bω {(1 + \omega )^7} = A + B\omega \,(1+ω)7=A+Bω. Then (A,B)(A,B)(A,B) equals :
  1. A
    (1 ,1)
  2. B
    (1, 0)
  3. C
    (- 1 ,1)
  4. D
    (0 ,1)
View written solutionFree

Correct answer: A

  1. Use the property of cube roots of unity

If ω≠1\omega \neq 1ω=1 is a cube root of unity, then ω3=1and1+ω+ω2=0.\omega^3=1 \quad \text{and} \quad 1+\omega+\omega^2=0.ω3=1and1+ω+ω2=0.

From 1+ω+ω2=0,1+\omega+\omega^2=0,1+ω+ω2=0, we get 1+ω=−ω2.1+\omega=-\omega^2.1+ω=−ω2.

  1. Compute (1+ω)7(1+\omega)^7(1+ω)7

Using 1+ω=−ω21+\omega=-\omega^21+ω=−ω2,

(1+ω)7=(−ω2)7=−ω14.(1+\omega)^7 = (-\omega^2)^7 = -\omega^{14}.(1+ω)7=(−ω2)7=−ω14.

Now reduce the power using ω3=1\omega^3=1ω3=1: 14≡2(mod3).14 \equiv 2 \pmod{3}.14≡2(mod3). So,

ω14=ω2.\omega^{14}=\omega^2.ω14=ω2.

Hence,

(1+ω)7=−ω2.(1+\omega)^7=-\omega^2.(1+ω)7=−ω2.
  1. Express in the form A+BωA+B\omegaA+Bω

Using 1+ω+ω2=0⇒ω2=−1−ω,1+\omega+\omega^2=0 \Rightarrow \omega^2=-1-\omega,1+ω+ω2=0⇒ω2=−1−ω, we get

−ω2=1+ω.-\omega^2 = 1+\omega.−ω2=1+ω.

Thus,

(1+ω)7=1+ω.(1+\omega)^7 = 1+\omega.(1+ω)7=1+ω.

Comparing with A+Bω,A+B\omega,A+Bω, we obtain

  1. Check options
  • A: (1,1)(1,1)(1,1) ✅
  • B: (1,0)(1,0)(1,0) ❌
  • C: (−1,1)(-1,1)(−1,1) ❌
  • D: (0,1)(0,1)(0,1) ❌

Therefore, the correct answer is Option A.

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