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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 \[(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )........\] \[(\cos n\theta +i\sin n\theta )=1\], then the value of \[\theta \] is[Karnataka CET 1992; Kurukshetra CEE 2002]

    A) \[4m\pi \]

    B) \[\frac{2m\pi }{n(n+1)}\]

    C) \[\frac{4m\pi }{n(n+1)}\]

    D) \[\frac{m\pi }{n(n+1)}\]

    Correct Answer: C

    Solution :

    We have \[(\cos \theta +i\sin \theta )(\cos 2\theta +i\sin 2\theta )\]   ......\[(\cos n\theta +i\sin n\theta )=1\] Þ \[\cos (\theta +2\theta +3\theta +...+n\theta )+i\sin (\theta +2\theta +.+n\theta )=1\] Þ  \[\cos \left( \frac{n(n+1)}{2}\theta  \right)+i\sin \left( \frac{n(n+1)}{2}\theta  \right)=1\] \[\cos \left( \frac{n\,(n+1)}{2}\theta  \right)\,=\,1\text{ and }\sin \left( \frac{n(n+1)}{2}\theta  \right)\,=0\] Þ \[\frac{n(n+1)}{2}\theta =2m\pi \Rightarrow \theta =\frac{4m\pi }{n(n+1)},\]where \[m\in I.\]


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