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 \[z=1+i,\] then the multiplicative inverse of z2 is (where i = \[\sqrt{-1}\]) [Karnataka CET 1999]

    A) 2 si

    B) 1 - i

    C) - i/2

    D) i/2

    Correct Answer: C

    Solution :

    Given \[z=1+i\] and \[i=\sqrt{-1}.\] Squaring both sides, we get \[{{z}^{2}}={{(1+i)}^{2}}=1+2i+{{i}^{2}}=1+2i-1\] or \[{{z}^{2}}=2i.\] Since it is multiplicative identity, therefore multiplicative inverse of \[{{z}^{2}}=\frac{1}{2i}\times \frac{i}{i}=\frac{i}{2{{i}^{2}}}=-\frac{i}{2}.\]


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