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

  • question_answer \[\left( \frac{1}{1-2i}+\frac{3}{1+i} \right)\,\,\left( \frac{3+4i}{2-4i} \right)=\] [Roorkee 1979; RPET 1999; Pb. CET 2003]

    A) \[\frac{1}{2}+\frac{9}{2}i\]

    B) \[\frac{1}{2}-\frac{9}{2}i\]

    C) \[\frac{1}{4}-\frac{9}{4}i\]

    D) \[\frac{1}{4}+\frac{9}{4}i\]

    Correct Answer: D

    Solution :

    \[\left( \frac{1}{1-2i}+\frac{3}{1+i} \right)\,\,\left( \frac{3+4i}{2-4i} \right)\] \[=\left[ \frac{1+2i}{{{1}^{2}}+{{2}^{2}}}+\frac{3-3i}{{{1}^{2}}+{{1}^{2}}} \right]\,\left[ \frac{6-16+12i+8i}{{{2}^{2}}+{{4}^{2}}} \right]\] \[=\left( \frac{2+4i+15-15i}{10} \right)\,\,\left( \frac{-1+2i}{2} \right)\] \[=\frac{(17-11i)(-1+2i)}{20}=\frac{5+45i}{20}=\frac{1}{4}+\frac{9}{4}i\].


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