Question

Reduce $(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)$ to the standard form.

Answer 1

Given data:
$z=(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)$
Taking L.C.M. and simplify
$=\left[\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}\right]\left[\frac{3-4i}{5+i}\right]$
$=\left[\frac{1+i-2+8i}{1+i-4i-4i^2}\right]\left[\frac{3-4i}{5+i}\right][\because i^2=-1]$
$=\left[\frac{-1+9i}{5-3i}\right]\left[\frac{3-4i}{5+i}\right]$
Multiply two or more complex number
$\left[\frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}\right]$
$=\frac{33+31i}{28-10i}$
$=\frac{33+31i}{2(14-5i)}$
On multiplying numerator and denominator by $(14+5i)$
$z=\frac{(33+31i)}{2(14-5i)}\times \frac{14+5i}{14+5i}$
$=\frac{462+165i+434i+155(i{)}^2}{2\left[\right(14{)}^2-\left(5i{)}^2\right]}$
$=\frac{307+599i}{2(196-25i^2)}$
$=\frac{307+599i}{2(196+25)}$
$=\frac{307+599i}{442}$
$=\frac{307}{442}+\frac{599}{442}i$