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

  • question_answer
    \[{{\left( \frac{1+i}{1-i} \right)}^{2}}+{{\left( \frac{1-i}{1+i} \right)}^{2}}\]is equal to

    A) \[2i\]

    B) \[-2i\]

    C) \[-2\]

    D) \[2\]

    Correct Answer: C

    Solution :

    \[{{\left( \frac{1+i}{1-i} \right)}^{2}}+{{\left( \frac{1-i}{1+i} \right)}^{2}}=\frac{2i}{-2i}+\left( \frac{-2i}{2i} \right)=-2\]


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