CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    If\[i=\sqrt{-1},\]then \[4+5{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{365}}\]is equal to:

    A)  \[1-i\sqrt{3}\]  

    B)         \[-1+i\sqrt{3}\]

    C)  \[i\sqrt{3}\]                      

    D)         \[-i\sqrt{3}\]

    E)  \[1+i\sqrt{3}\]

    Correct Answer: C

    Solution :

    We knows\[-\frac{1}{2}+\frac{i\sqrt{3}}{2}=\omega \] \[\therefore \]  \[4+5{{(\omega )}^{334}}+3{{(\omega )}^{365}}\] \[=4+5{{({{\omega }^{3}})}^{111}}.{{\omega }^{1}}+3{{({{\omega }^{3}})}^{121}}.{{\omega }^{2}}\] \[=4+5\omega +3{{\omega }^{2}}\] \[=4+5\left( -\frac{1}{2}+\frac{i\sqrt{3}}{\sqrt{2}} \right)+3\left( -\frac{1}{2}-\frac{i\sqrt{3}}{2} \right)\] \[=4-\frac{5}{2}+\frac{5i\sqrt{3}}{2}-\frac{3}{2}-\frac{3i\sqrt{3}}{2}\] \[=i\sqrt{3}\]


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