Question

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

Answer 1

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

which is the required standard form.