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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank De Moivre's theorem and Roots of unity

  • question_answer
    If \[\omega \] is a complex cube root of unity, then the value of \[{{\omega }^{99}}+{{\omega }^{100}}+{{\omega }^{101}}\] is [Pb. CET 2004]

    A) 1

    B) - 1

    C) 3

    D) 0

    Correct Answer: D

    Solution :

    \[{{\omega }^{99}}+{{\omega }^{1\omega }}+{{\omega }^{101}}\] = \[{{\omega }^{99}}[1+\omega +{{\omega }^{2}}]\] [Since \[1+\omega +{{\omega }^{2}}=0,\,{{\omega }^{3}}=1\]] = 0.


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