Reduce to the standard form.


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


$\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)=\left[\frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)}\right]\left[\frac{3-4 i}{5+i}\right]$

$=\left[\frac{1+i-2+8 i}{1+i-4 i-4 i^{2}}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{-1+9 i}{5-3 i}\right]\left[\frac{3-4 i}{5+i}\right]$

$=\left[\frac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}\right]=\frac{33+31 i}{28-10 i}=\frac{33+31 i}{2(14-5 i)}$

$=\frac{(33+31 i)}{2(14-5 i)} \times \frac{(14+5 i)}{(14+5 i)}$  [On multiplying numerator and denominator by $(14+5 i)]$

$=\frac{462+165 i+434 i+155 i^{2}}{2\left[(14)^{2}-(5 i)^{2}\right]}=\frac{307+599 i}{2\left(196-25 i^{2}\right)}$

$=\frac{307+599 i}{2(221)}=\frac{307+599 i}{442}=\frac{307}{442}+\frac{599 i}{442}$

This is the required standard form.

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