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BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    The minimum value of n for\[{{\left( \frac{1+i}{1-i} \right)}^{n}}=1,\] where \[i=\sqrt{-1}\] is:

    A)  2

    B)                                         4

    C)  6                            

    D)  3

    Correct Answer: B

    Solution :

    \[{{\left( \frac{1+i}{1-i} \right)}^{n}}={{\left( \frac{1+i}{1-i}\times \frac{1+i}{1+i} \right)}^{n}}\] \[={{\left( \frac{1-1+2i}{{{1}^{2}}+{{1}^{2}}} \right)}^{n}}\]                 \[\Rightarrow \]               \[1={{(i)}^{n}}\]                 \[\Rightarrow \]\[n\]is a multiple of 4. \[\therefore \]The minimum value of n is 4


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