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Complex Numbers question

2005 · Shift 0 · Q70
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Complex Numbers question

2005 · Shift 0 · Q70

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If the cube roots of unity are 1, ω , ω2\omega \,,\,{\omega ^2}ω,ω2 then the roots of the equation (x−1)3{(x - 1)^3}(x−1)3 + 8 = 0, are :
  1. A
    −1,−1+2  ω,−1−2  ω2- 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}−1,−1+2ω,−1−2ω2
  2. B
    −1,−1,−1- 1, - 1, - 1−1,−1,−1
  3. C
    −1,1−2ω,1−2ω2- 1,1 - 2\omega ,1 - 2{\omega ^2}−1,1−2ω,1−2ω2
  4. D
    −1,1+2ω,1+2ω2- 1,1 + 2\omega ,1 + 2{\omega ^2}−1,1+2ω,1+2ω2
View written solutionFree

Correct answer: C

  1. We need to solve
(x−1)3+8=0.(x-1)^3+8=0.(x−1)3+8=0.

So,

(x−1)3=−8.(x-1)^3=-8.(x−1)3=−8.
  1. Write −8-8−8 in a form involving cube roots of unity. Since
−8=8(−1)=8ei(π+2kπ),-8=8(-1)=8e^{i(\pi+2k\pi)},−8=8(−1)=8ei(π+2kπ),

its cube roots are

2ei(π+2kπ)/3,k=0,1,2.2e^{i(\pi+2k\pi)/3},\quad k=0,1,2.2ei(π+2kπ)/3,k=0,1,2.

Equivalently, the cube roots of −8-8−8 are

−2, −2ω, −2ω2-2,\, -2\omega,\, -2\omega^2−2,−2ω,−2ω2

up to ordering/sign simplification.

A cleaner way is:

(x−1)3+23=0(x-1)^3+2^3=0(x−1)3+23=0

which is of the form

a3+b3=0  ⟹  a=−b, −bω, −bω2.a^3+b^3=0 \implies a=-b,\,-b\omega,\,-b\omega^2.a3+b3=0⟹a=−b,−bω,−bω2.

Hence,

x−1=−2, −2ω, −2ω2.x-1=-2,\, -2\omega,\, -2\omega^2.x−1=−2,−2ω,−2ω2.
  1. Therefore,
x=1−2, 1−2ω, 1−2ω2.x=1-2,\, 1-2\omega,\, 1-2\omega^2.x=1−2,1−2ω,1−2ω2.

So the roots are

−1,1−2ω,1−2ω2.-1,\quad 1-2\omega,\quad 1-2\omega^2.−1,1−2ω,1−2ω2.
  1. Compare with the options:
  • Option A: incorrect
  • Option B: incorrect
  • Option C: −1,  1−2ω,  1−2ω2-1,\;1-2\omega,\;1-2\omega^2−1,1−2ω,1−2ω2 correct
  • Option D: incorrect

Hence the correct option is C.

  1. Comparison with stored answer: Stored correct answer = C, which matches our derived answer.
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