s6.tex

Applying the rules for the multiplication and division of
  complex numbers yields:

$
\begin{array}[h!]{lll}
{\rm (a)}\hskip5pt (5-i)(2+3\,i)=10+15\,i-2\,i+3=13+13\,i\, .\\
\noalign{\vskip12pt}
{\rm (b)}\hskip5pt (3-4\,i)(3+4\,i)=9+12\,i-12\,i+16=25\, .\\
\noalign{\vskip12pt}
{\rm (c)}\hskip5pt (1+2\,i)^2=(1+2\,i)(1+2\,i)=1+2\,i+2\,i-4=-3+4\,i\, .\\
\noalign{\vskip12pt}
{\rm (d)}\hskip5pt \displaystyle{{10\over4-2\,i}={10\over4-2\,i}\times{4+2\,i\over4+2\,i}=
{40+20\,i\over16+8\,i-8\,i+4}={40+20\,i\over20}=2+i\, .}\\
\noalign{\vskip12pt}
{\rm (e)}\hskip5pt \displaystyle{{3-i\over4+3\,i}={3-i\over4+3\,i}\times{4-3\,i\over4-3\,i}=
{12-9\,i-4\,i-3\over16-12\,i+12\,i+9}={9-13\,i\over25}={9\over25}-i\,{13\over25}\, .}\\
\noalign{\vskip12pt}
{\rm (f)}\hskip5pt \displaystyle{{1\over i}={1\over i}\times{-i\over-i}=-i\, .}
\end{array}
$