# Reduce $\;(\large\frac{1}{1-4i}- \large\frac{2}{1+i})\;(\large\frac{3 -4i}{5+i})\;$ to the standard form

$(a)\;\large\frac{27}{442}+ \large\frac{599}{442} i\qquad(b)\;\large\frac{599}{442}+ \large\frac{307}{442} i\qquad(c)\;\large\frac{207}{442}+ \large\frac{599}{442} i\qquad(d)\;\large\frac{307}{442}+ \large\frac{599}{442} i$

Answer : $\;\large\frac{307}{442}+ \large\frac{599}{442} i$
Explanation :
$\;(\large\frac{1}{1-4i}- \large\frac{2}{1+i})\;(\large\frac{3 -4i}{5+i})\; = \large\frac{(1+i) -2(1-4i)}{(1-4i)(1+i)}$
$= [\large\frac{1+i-2+8i}{1+i-4i-4i^{2}}]\;(\large\frac{3 -4i}{5+i})$
$= [\large\frac{-1+9i}{1-3i-4(-1)}]\;(\large\frac{3 -4i}{5+i}) \qquad [i^{2}=-1]$
$= [\large\frac{-1+9i}{5-3i}] \;(\large\frac{3 -4i}{5+i})$
$= \large\frac{-3+4i+27i-36i^{2}}{25+5i-15i-3i^{2}}$
$= \large\frac{33 +31i}{28-10i}$
$= \large\frac{33 +31i}{2(14-5i)}$
$= \large\frac{33 +31i}{2(14-5i)} \times \large\frac{(14+5i)}{(14+5i)}\;$ [on multiplying numerator and denominator by $\;14+5i$ ]
$= \large\frac{462 + 165i+434i+155i^{2}}{2[(14)^2-(5i)^{2}]}$
$= \large\frac{307+599i}{2(196-25i^{2})}$
$= \large\frac{307+599i}{2(221)}$
$= \large\frac{307+599i}{442}$
$=\large\frac{307}{442}+ \large\frac{599}{442} i$