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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 \[1,\omega ,{{\omega }^{2}}\] are the cube roots of unity, then\[\Delta =\left| \,\begin{matrix}    1\,\,\,\, & {{\omega }^{n}} & {{\omega }^{2n}}  \\    {{\omega }^{n}}\,\, & \,\,\,{{\omega }^{2n}}\,\, & 1  \\    {{\omega }^{2n}}\, & 1\,\, & {{\omega }^{n}}  \\ \end{matrix} \right|\]= [AIEEE 2003]

    A) 0

    B) 1

    C) \[\omega \]

    D) \[{{\omega }^{2}}\]

    Correct Answer: A

    Solution :

    \[\Delta =\,({{\omega }^{3n}}-1)+{{\omega }^{n}}({{\omega }^{2n}}-{{\omega }^{2n}})+{{\omega }^{2n}}({{\omega }^{n}}-{{\omega }^{4n}})\] \[\Delta \,=(1-1)+0+{{\omega }^{2n}}[{{\omega }^{n}}-{{({{\omega }^{3}})}^{n}}{{\omega }^{n}}]\] \[\Delta =0+0+0=0\].


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