Evaluate [1 1-4 i -2 1+i ] [3-4 i 5+i ] to the standard form.

Question

Evaluate $\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]$ to the standard form.

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Answer

${\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{1+i-2+8 i}{(1-4 i)(1+i)}\right]\left[\frac{3-4 i}{5+i}\right]}$
$=\left[\frac{-1+9 i}{1+i-4 i-4 i^2}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{-1+9 i}{5-3 i}\right]\left[\frac{3-4 i}{5+i}\right]$
$=\frac{-3+4 i+27 i-36 i^2}{25+5 i-15 i-3 i^2}=\frac{33+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i}$
$=\frac{924+330 i+868 i+310 i^2}{(28)^2-(10 i)^2}=\frac{614+1198 i}{784+100}\left(\because i^2=-1\right)$
$=\frac{2(307+599 i)}{884}=\frac{307+599 i}{442}$

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