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JEE Advanced 2022 Paper 1 · Mathematics · Complex Numbers · hard

Fix the primitive seventh root of unity ω=cos (2π)/(7)+isin (2π)/(7) and define α=ω+ω^2+ω^4, β=ω^3+ω^5+ω^6.

One or more options correct

Fix the primitive seventh root of unity ω=cos⁡2π7+isin⁡2π7\omega=\cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7} and define α=ω+ω2+ω4\alpha=\omega+\omega^2+\omega^4, β=ω3+ω5+ω6\beta=\omega^3+\omega^5+\omega^6. Choose the correct statement(s).
  1. A. α\alpha and β\beta are the roots of x2+x+2=0x^2+x+2=0.
  2. B. Im⁡(α)<0\operatorname{Im}(\alpha)<0.
  3. C. α2+β2=3\alpha^2+\beta^2=3.
  4. D. ∣α−β∣=27|\alpha-\beta|=2\sqrt7.
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Answer: A

Since 1+ω+⋯+ω6=01+\omega+\dots+\omega^6=0, α+β=−1\alpha+\beta=-1. Expanding αβ\alpha\beta gives nine terms: three equal to ω7=1\omega^7=1 and the six powers ωk\omega^k, k=1,…,6k=1,\dots,6, each once, so αβ=3+(−1)=2\alpha\beta=3+(-1)=2. (A) true: roots of x2+x+2=0x^2+x+2=0, i.e. −1±i72\frac{-1\pm i\sqrt7}2. (B) Im⁡α=sin⁡2π7+sin⁡4π7+sin⁡8π7>0\operatorname{Im}\alpha=\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7>0 (numerically 72\frac{\sqrt7}2), so false. (C) α2+β2=1−4=−3\alpha^2+\beta^2=1-4=-3: false. (D) (α−β)2=(α+β)2−4αβ=1−8=−7(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=1-8=-7, so ∣α−β∣=7≠27|\alpha-\beta|=\sqrt7\ne2\sqrt7: false.
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