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Complex Numbers question

2021 · 25 Feb · Shift 2 · Q33
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  5. /2021 · 25 Feb · Shift 2 · Q33

Complex Numbers question

2021 · 25 Feb · Shift 2 · Q33

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If α\alphaα, β∈\beta\inβ∈ R are such that 1 −-− 2i (here i2 =−-− 1) is a root of z2 + α\alphaα z + β\betaβ = 0, then (α−β\alpha-\betaα−β) is equal to :
  1. A
    −-− 7
  2. B
    7
  3. C
    3
  4. D
    −-− 3
View written solutionFree

Correct answer: A

  1. Since the coefficients α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R, the quadratic z2+αz+β=0z^2 + \alpha z + \beta = 0z2+αz+β=0 has real coefficients.

  2. If 1−2i1-2i1−2i is a root, then its complex conjugate 1+2i1+2i1+2i must also be a root.

  3. So the quadratic can be written as (z−(1−2i))(z−(1+2i))=0.(z-(1-2i))(z-(1+2i)) = 0.(z−(1−2i))(z−(1+2i))=0.

  4. Expand: [ (z-1+2i)(z-1-2i) = (z-1)^2 - (2i)^2 = (z-1)^2 + 4 ] because i2=−1i^2=-1i2=−1, so (2i)2=−4(2i)^2 = -4(2i)2=−4.

  5. Now, [ (z-1)^2 + 4 = z^2 - 2z + 1 + 4 = z^2 - 2z + 5. ]

  6. Comparing with z2+αz+β,z^2 + \alpha z + \beta,z2+αz+β, we get α=−2,β=5.\alpha = -2, \qquad \beta = 5.α=−2,β=5.

  7. Therefore, α−β=−2−5=−7.\alpha - \beta = -2 - 5 = -7.α−β=−2−5=−7.

  8. Hence the correct option is: A: −7\boxed{\text{A: } -7}A: −7​

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