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

  • question_answer
    \[{{\left( \frac{\sqrt{3}+i}{2} \right)}^{6}}+{{\left( \frac{i-\sqrt{3}}{2} \right)}^{6}}\]is equal to [RPET 1997]

    A) \[-2\]

    B) 0

    C) 2

    D) 1

    Correct Answer: A

    Solution :

      \[{{\left( \frac{\sqrt{3}+i}{2} \right)}^{6}}+{{\left( \frac{i-\sqrt{3}}{2} \right)}^{6}}={{\left( \frac{-1+\sqrt{3}i}{2i} \right)}^{6}}+{{\left( \frac{-1-\sqrt{3}i}{2i} \right)}^{6}}\] \[=\frac{1}{{{i}^{6}}}[{{(\omega )}^{6}}+{{({{\omega }^{2}})}^{6}}]=-[{{({{\omega }^{3}})}^{2}}+{{({{\omega }^{3}})}^{4}}]\]\[\left( \because \,\,\,\omega =\frac{-1+\sqrt{3}i}{2},{{\omega }^{2}}=\frac{-1-\sqrt{3}i}{2} \right)\] \[=-(1+1)=-2\].


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