Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    If \[{{E}_{2}}\]then \[{{r}_{1}}\]is equal to

    A) 1                                             

    B)  ? 1

    C) i                                              

    D) -i

    Correct Answer: C

    Solution :

    We have, \[{{x}_{n}}=\cos \left( \frac{\pi }{{{3}^{n}}} \right)+i\sin \left( \frac{\pi }{{{3}^{n}}} \right)\] \[\therefore \]\[{{x}_{1}}.{{x}_{2}}.{{x}_{3}}...\] \[=\cos \left( \frac{\pi }{3}+\frac{\pi }{{{3}^{2}}}+... \right)+i\sin \left( \frac{\pi }{3}+\frac{\pi }{{{3}^{2}}}+... \right)\] \[=\cos \left( \frac{\pi /3}{1-\frac{1}{3}} \right)+i\sin \left( \frac{\pi /3}{1+\frac{1}{3}} \right)\] \[=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=0+i=i\]


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