Question

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

Answer 1

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

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

we know that $i^2=\sqrt{-1}\times \sqrt{-1}=-1$

so, $[\frac{1+4i}{1-(4i{)}^2}-\frac{2(1-i)}{1-i^2}]\left[\frac{15-3i-20i+4i^2}{5-i^2}\right]$

$\Rightarrow [\frac{1+4i}{17}-\frac{1-i}{1}]\left[\frac{11-23i}{6}\right]$

$\Rightarrow [\frac{1}{17}+\frac{4}{17}i-1+i][\frac{11}{6}-\frac{23}{6}i]$

$\Rightarrow [\frac{-16}{17}+\frac{21}{17}i][\frac{11}{6}-\frac{23}{6}i]$

$\Rightarrow \frac{16}{17}\times \frac{11}{6}+\frac{16}{17}\times \frac{23}{6}i+\frac{21}{17}i\times \frac{11}{6}-\frac{21}{17}\times \frac{23}{6}\left(i^2\right)$

$\Rightarrow \frac{-88}{51}+\frac{184}{51}i+\frac{231}{102}i+\frac{483}{102}$

$\Rightarrow \frac{307}{102}+\frac{599}{102}i$

$\Rightarrow \frac{1}{102}[599i+307]$ Answer