JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer
    The smallest positive integer \[n\]for which \[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\]is  [Karnataka CET 2004]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: B

    Solution :

    We have \[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\] \[\Rightarrow {{\left( \frac{1+i}{1-i} \right)}^{2n}}=1\]\[\Rightarrow {{(i)}^{2n}}=1\]\[\Rightarrow {{(i)}^{2n}}={{(-1)}^{2}}\] \[\Rightarrow {{(i)}^{2n}}={{({{i}^{2}})}^{2}}\]\[\Rightarrow {{(i)}^{2n}}={{(i)}^{4}}\]\[\Rightarrow 2n=4\]\[\Rightarrow n=2\].

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