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High power of a complex number using polar form (De Moivre)

This type of question, asked 5 times: 2021 Paper 2, 2021 Paper 19, 29 Jul 2022, 24 Jan 2023, 23 Jan 2026 S2. All 6 questions below, oldest first.

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  1. Question 1

    2021 Paper 2

    86
    Let i=−1i=\sqrt{- 1}. If (−1+i3)21(1−i)24+(1+i3)21(1+i)24=k\frac{( - 1 +i\sqrt{3})^{21}}{(1 -i)^{24}}+\frac{(1 +i\sqrt{3})^{21}}{(1 +i)^{24}}=k, and n=[∣k∣]n = \lbrack|k|\rbrack be the greatest integral part of ∣k∣|k|. Then ∑j=0n+5(j+5)2−∑j=0n+5(j+5)\sum_{j= 0}^{n+ 5} (j+ 5)^{2}-\sum_{j= 0}^{n+ 5} (j+ 5) is equal to
  2. Question 2

    2021 Paper 19

    74
    If (3+i)100=299(p+iq)(\sqrt{3}+ i)^{100}=2^{99}(p + iq), then pp and qq are roots of the equation:

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  3. Question 3

    29 Jul 2022

    62
    If z=2+3iz = 2 + 3i, then z5+(Z‾)5z^{5}+ (\overline{Z})^{5} is equal to:

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  4. Question 4

    24 Jan 2023

    170
    The value of (1+sin⁡2π9+icos2π91+sin⁡2π9−icos2π9)3\left( \frac{1 + \sin\frac{2\pi}{9}+ icos\frac{2\pi}{9}}{1 + \sin\frac{2\pi}{9}- icos\frac{2\pi}{9}} \right)^{3} is

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  5. Question 5

    24 Jan 2023

    65
    Let p,q∈Rp,q \in R and (1−3i)200=2199(p+iq),i=−1(1 -\sqrt{3}i)^{200}=2^{199}(p + iq),i =\sqrt{- 1} Then p+q+q2p + q +q^{2} and p−q+q2p - q +q^{2} are roots of the equation

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  6. Question 6

    23 Jan 2026 S2

    67
    If z=32+i2,i=−1z =\frac{\sqrt{3}}{2}+\frac{i}{2},i =\sqrt{- 1}, then (z201−i)8\left( z^{201}- i \right)^{8} is equal to

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