Question

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

• A. $\frac{1}{2}+\left(\frac{9}{2}\right)i$
• B. $\frac{1}{2}-\left(\frac{9}{2}\right)i$
• C. $\frac{1}{4}-\left(\frac{9}{4}\right)i$
• D. $\frac{1}{4}+\left(\frac{9}{4}\right)i$

Answer 1

The correct option is D

$\frac{1}{4}+\left(\frac{9}{4}\right)i$

Explanation for the correct option:

Step 1. Simplify the given expression:

$\left(\frac{1}{(1-2i)}+\frac{3}{(1+i)}\right)\left(\frac{3+4i}{2-4i}\right)$ $=\left[\frac{1+i+3-6i}{(1-2i)(1+i)}\right]\left[\frac{3+4i}{2-4i}\right]$

$=\left[\frac{4-5i}{3-i}\right]\left[\frac{3+4i}{2-4i}\right]$

$=\left[\frac{12+16i-15i+20}{6-12i-2i-4}\right]$ $\left[\mathbf{\because }{\mathit{i}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{1}\right]$

$=\left[\frac{32+i}{2-14i}\right]$

Step 2. By Rationalizing denominator, we get

$\frac{32+i}{2-14i}\times \frac{2+14i}{2+14i}=\frac{64+448i+2i-14}{4+196}$

$$\begin{gathered} =\frac{50+450i}{200} \\ =\frac{\mathbf{1}}{\mathbf{4}}\mathbf{}\mathbf{+}\frac{\mathbf{9}}{\mathbf{4}}\mathit{i} \end{gathered}$$

Hence, option ‘D’ is Correct.