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 \[{{\left( \frac{1+i}{1-i} \right)}^{m}}=1,\]then the least integral value of \[m\] is [IIT 1982; MNR 1984; UPSEAT 2001; MP PET 2002]

    A) 2

    B) 4

    C) 8

    D) None of these

    Correct Answer: B

    Solution :

    \[\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{{{(1+i)}^{2}}}{2}=\frac{2i}{2}=i\] \[\therefore \]\[{{\left( \frac{1+i}{1-i} \right)}^{m}}={{i}^{m}}=1\] (as given) So the least value of \[m=4\]\[\{\because {{i}^{4}}=1\}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner