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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 for positive integral value of\[n\], the product of \[\omega .{{\omega }^{2}}.{{\omega }^{3}}........{{\omega }^{n}}\], will be [Roorkee 1991]

    A) \[\frac{1-i\sqrt{3}}{2}\]

    B) \[-\frac{1-i\sqrt{3}}{2}\]

    C) 1

    D) (b) and (c) both

    Correct Answer: D

    Solution :

    The product is given by \[\omega .{{\omega }^{2}}.{{\omega }^{3}}.....{{\omega }^{n}}={{\omega }^{1+2+3+......+n}}={{\omega }^{n(n+1)/2}}\] On putting \[n=1,2,3,.....,\]we get \[={{\omega }^{1(1+1)/2}}=\omega ,{{\omega }^{2(2+1)/2}}={{\omega }^{3}}=1,.....{{\omega }^{4(5)/2}}={{\omega }^{10}}=\omega \] Hence it gives the values 1 and \[\omega \] only.


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