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

  • question_answer
    We express \[\frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}}\]  in the form of \[x+iy\], we get [Karnataka CET 2001]

    A) \[\cos 49\theta -i\,\sin 49\theta \]

    B) \[\cos 23\theta -i\,\sin 23\theta \]

    C) \[\cos 49\theta +i\,\sin 49\theta \]

    D) \[\cos 21\theta +i\,\sin 21\theta \]

    Correct Answer: A

    Solution :

    \[\frac{{{(\cos 2\theta -i\sin 2\theta )}^{4}}\,{{(\cos 4\theta +i\sin 4\theta )}^{-5}}}{{{(\cos 3\theta +i\sin 3\theta )}^{-2}}{{(\cos 3\theta -i\sin 3\theta )}^{-9}}}\] \[=\frac{{{[{{(\cos \theta +i\sin \theta )}^{-2}}]}^{4}}{{[{{(\cos \theta +i\sin \theta )}^{4}}]}^{-5}}}{{{[{{(\cos \theta +i\sin \theta )}^{3}}]}^{-2}}\,{{[{{(\cos \theta +i\sin \theta )}^{-3}}]}^{-9}}}\] \[=\frac{{{(\cos \theta +i\sin \theta )}^{-8}}{{(\cos \theta +i\sin \theta )}^{-20}}}{{{(\cos \theta +i\sin \theta )}^{-6}}\,{{(\cos \theta +i\sin \theta )}^{27}}}\] \[={{(\cos \theta +i\sin \theta )}^{-8-20+6-27}}={{(\cos \theta +i\sin \theta )}^{-49}}\] \[=\cos 49\theta -i\sin 49\theta .\]


You need to login to perform this action.
You will be redirected in 3 sec spinner