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

  • question_answer
    If  \[{{x}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right)\], then\[{{x}_{1}}.{{x}_{2}}......\infty \]is [RPET 1990, 2000; BIT Mesra 1996; Karnataka CET 2000]

    A) \[-3\]

    B) \[-2\]

    C) \[-1\]

    D)   0

    Correct Answer: C

    Solution :

    \[{{x}_{1}},{{x}_{2}},{{x}_{3}}.....\]upto\[\infty =\]\[\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\,\,\left( \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right)\] upto .....\[\infty \] \[=\cos \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+..... \right)\]\[+i\sin \left( \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+..... \right)\] \[=\cos \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)+i\sin \left( \frac{\frac{\pi }{2}}{1-\frac{1}{2}} \right)\]\[=\cos \pi +i\sin \pi =-1\].


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