Q. $\left( \frac{1}{1 + 2i} + \frac{3}{1 + i} \right)\left( \frac{3 + 4i}{2 - 4i} \right)$ is equal to

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) $\frac{1}{4} + \frac{9}{4}i$
Explanation

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

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.