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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    If \[i=\sqrt{-1},\]then \[1+{{i}^{2}}+{{i}^{3}}-{{i}^{6}}+{{i}^{8}}\] is equal to:

    A) \[2-i\]                                   

    B)  1                            

    C)  3                            

    D)         \[-1\]

    Correct Answer: A

    Solution :

    \[1+{{i}^{2}}+{{i}^{3}}-{{i}^{6}}+{{i}^{8}}\] \[=1+(-1)+(-i)-(-1)+1\] \[[\because \,i=\sqrt{-1}]\]                                 \[=2-i\]


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