CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    If\[{{z}_{r}}=\cos \left( \frac{\pi }{{{2}^{r}}} \right)+i\sin \left( \frac{\pi }{{{2}^{r}}} \right),\]then\[{{z}_{1}}.{{z}_{2}}.{{z}_{3}}\]upto \[\infty \]equals:

    A)  \[-3\]                   

    B)         \[-2\]

    C)  1                            

    D)         0

    E)  1                         

    Correct Answer: E

    Solution :

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


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