Question

$\left(\frac{1}{1-2i} + \frac{3}{1+i}\right) \left(\frac{3+4i}{2-4i}\right)$ 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. $\frac{1}{4}+\frac{9}{4}i$

Answer 1

Answer: (D)

Let $z = \left(\frac{1}{1-2i} + \frac{2}{1+i}\right) \left(\frac{3+4i}{2-4i}\right) $
$= \left[\frac{1+i+3-6i}{\left(1-2i\right)\left(1+i\right)}\right] \left[\frac{3+4i}{2-4i}\right]$
$= \left[\frac{4-5i}{3-i}\right]\left[\frac{3+4i}{2-4i}\right] = \left[\frac{32+i}{2-14i}\right]$
$= \frac{32+i}{2-14i}\times\frac{2+14i}{2+14i} = \frac{64+448i+2i-14}{4+196}$
$= \frac{50+450i}{200} = \frac{1}{4} + \frac{9}{4}i$