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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If\[{{(\sqrt{8}+i)}^{50}}={{3}^{49}}(a+ib),\]then\[{{a}^{2}}+{{b}^{2}}\]is:

    A)  3                            

    B)         8

    C)  9                            

    D)         \[\sqrt{8}\]

    E)  4

    Correct Answer: C

    Solution :

    Let\[\sqrt{8}+i=r(\cos \theta +i\sin \theta )\] Here,\[r=3\]and \[\theta ={{\tan }^{-1}}\left( \frac{1}{2\sqrt{2}} \right)\] Now, \[{{(\sqrt{8}+i)}^{50}}={{3}^{50}}{{(\cos \theta +i\sin \theta )}^{50}}\] But \[{{(\sqrt{8}+i)}^{50}}={{3}^{48}}(a+ib)\] \[\therefore \]\[a=3\text{ }cos\theta \]and\[b=3\text{ }sin\theta \] So, \[{{a}^{2}}+{{b}^{2}}={{(3\text{ }cos\theta )}^{2}}+{{(3\text{ }sin\theta )}^{2}}\] \[=9\]


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