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

  • question_answer If  \[n\] is a positive integer,  then \[{{\left( \frac{1+i}{1-i} \right)}^{4n+1}}\]=

    A) 1

    B) - 1

    C) \[i\]

    D) \[-i\]

    Correct Answer: C

    Solution :

    Since \[\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=i\] Therefore \[{{\left( \frac{1+i}{1-i} \right)}^{4n+1}}={{i}^{4n+1}}=i{{i}^{4n}}=i\,\,\,\,\,(\because {{i}^{4n}}=1)\]


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