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Complex Numbers question

2022 · 29 Jun · Shift 1 · Q26
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  5. /2022 · 29 Jun · Shift 1 · Q26

Complex Numbers question

2022 · 29 Jun · Shift 1 · Q26

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let α\alphaα and β\betaβ be the roots of the equation x2 + (2i −-− 1) = 0. Then, the value of |α\alphaα 8 + β\betaβ 8| is equal to :
  1. A
    50
  2. B
    250
  3. C
    1250
  4. D
    1500
View written solutionFree

Correct answer: A

  1. Interpret the equation

The given equation is x2+(2i−1)=0x^2+(2i-1)=0x2+(2i−1)=0 so x2=1−2i.x^2=1-2i.x2=1−2i.

Hence the roots α,β\alpha,\betaα,β satisfy α2=1−2i,β2=1−2i.\alpha^2=1-2i, \qquad \beta^2=1-2i.α2=1−2i,β2=1−2i. Since they are roots of a quadratic of the form x2−a=0x^2-a=0x2−a=0, the two roots are opposites: β=−α.\beta=-\alpha.β=−α.

  1. Relate α8\alpha^8α8 and β8\beta^8β8

Because 888 is even, β8=(−α)8=α8.\beta^8=(-\alpha)^8=\alpha^8.β8=(−α)8=α8. Therefore, α8+β8=2α8.\alpha^8+\beta^8=2\alpha^8.α8+β8=2α8. So, ∣α8+β8∣=2∣α8∣=2∣α∣8.|\alpha^8+\beta^8|=2|\alpha^8|=2|\alpha|^8.∣α8+β8∣=2∣α8∣=2∣α∣8.

  1. Find ∣α∣|\alpha|∣α∣

From α2=1−2i,\alpha^2=1-2i,α2=1−2i, we get ∣α∣2=∣1−2i∣.|\alpha|^2=|1-2i|.∣α∣2=∣1−2i∣. Now, ∣1−2i∣=12+(−2)2=5.|1-2i|=\sqrt{1^2+(-2)^2}=\sqrt{5}.∣1−2i∣=12+(−2)2​=5​. Thus, ∣α∣2=5.|\alpha|^2=\sqrt{5}.∣α∣2=5​. Squaring again, ∣α∣4=5,|\alpha|^4=5,∣α∣4=5, and hence, ∣α∣8=25.|\alpha|^8=25.∣α∣8=25.

  1. Compute the required value

Therefore, ∣α8+β8∣=2⋅25=50.|\alpha^8+\beta^8|=2\cdot 25=50.∣α8+β8∣=2⋅25=50.

  1. Check options
  • A: 505050 ✅
  • B: 250250250 ❌
  • C: 125012501250 ❌
  • D: 150015001500 ❌

So the correct option is A.

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