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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    The smallest positive integer n for which\[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\] is :

    A)  1                                            

    B)  2

    C)  3                                            

    D)  4

    Correct Answer: B

    Solution :

    Given that, \[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\] \[\Rightarrow \]               \[{{(1-1+2i)}^{n}}={{(1-1-2i)}^{n}}\]                                 \[{{2}^{n}}\,{{i}^{n}}={{2}^{n}}\,{{(-1)}^{n}}\,{{i}^{n}}\]                                                 \[1={{(-1)}^{n}}\] \[\therefore \] The smallest value of n is 2.


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