• # 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) $\frac{1}{4}+\frac{9}{4}i$

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