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Sum of high powers of the complex roots of a quadratic

This type of question, asked 4 times: 29 Jun 2022, 13 Apr 2023, 29 Jan 2024, 6 Apr 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    29 Jun 2022

    68
    Let α\alpha and β\beta be the roots of the equation x2+(2i−x^{2}+ (2i - 1)=01) = 0. Then, the value of ∣α8+β8∣\left| \alpha^{8}+\beta^{8} \right| is equal to:

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  2. Question 2

    13 Apr 2023

    154
    Let α,β\alpha,\beta be the roots of the equation x2−2x+2=0x^{2}-\sqrt{2}x + 2 = 0. Then α14+β14\alpha^{14}+\beta^{14} is equal to

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  3. Question 3

    29 Jan 2024

    171
    Let α,β\alpha,\beta be the roots of the equation x2−6x+3=0x^{2}-\sqrt{6}x+ 3 = 0 such that Im(α)>Im(β)Im(\alpha) > Im(\beta). Let a,ba,b be integers not divisible by 3 and nn be a natural number such that α99β+α98=3n(a+ib),i=−1\frac{\alpha^{99}}{\beta}+\alpha^{98}=3^{n}(a+ib),i=\sqrt{- 1}. Then n+a+bn + a + b is equal to
  4. Question 4

    29 Jan 2024

    83
    Let α,β\alpha,\beta be the roots of the equation x2−x+2=0x^{2}- x + 2 = 0 with Im(α)>Im(β)Im(\alpha) > Im(\beta). Then α6+α4+β4−5α2\alpha^{6}+\alpha^{4}+\beta^{4}- 5\alpha^{2} is equal to
  5. Question 5

    6 Apr 2026 S2

    53
    Let S={z∈C:z2+6iz−3=0}S=\left\{ z\in C:z^{2}+\sqrt{6}iz- 3 = 0 \right\}. Then ∑z∈Sz8\sum_{z\in S} z^{8} is equal to :

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