CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer If \[\omega \] is a complex cube root of unity, then \[\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}+\frac{c+a\omega +b{{\omega }^{2}}}{a+b\omega +c{{\omega }^{2}}}+\frac{b+c\omega +a{{\omega }^{2}}}{b+c{{\omega }^{4}}+a{{\omega }^{5}}}\] is equal to:

    A)  \[1\]                    

    B)         \[\omega \]                      

    C)  \[{{\omega }^{2}}\]                       

    D)         \[-1\]

    E)  0

    Correct Answer: E

    Solution :

    \[\frac{a+b\omega +c{{\omega }^{2}}}{c+a\omega +b{{\omega }^{2}}}+\frac{c+a\omega +b{{\omega }^{2}}}{a+b\omega +c{{\omega }^{2}}}+\frac{b+c\omega +a{{\omega }^{2}}}{b+c{{\omega }^{4}}+a{{\omega }^{5}}}\] \[=\frac{{{\omega }^{2}}(a+b\omega +c{{\omega }^{2}})}{(a+b\omega +c{{\omega }^{2}})}+\frac{\omega (a\omega +b{{\omega }^{2}}+c)}{(a\omega +b{{\omega }^{2}}+c)}\]                                                 \[+\frac{b+c\omega +a{{\omega }^{2}}}{b+c\omega +a{{\omega }^{2}}}\] \[={{\omega }^{2}}+\omega +1=0\]

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