s5.tex

For a complex number $a+i\,b$, in which $a$ and $b$ are real, the real and imaginary parts
are given by ${\rm Re}(a+i\,b)=a$ and ${\rm Im}(a+i\,b)=b$, respectively.  Thus,

$
\begin{array}[h!]{l}
{\rm (a)}\hskip5pt {\rm Re}(8+3\,i)=8\, ,\qquad{\rm Im}(8+3\,i)=3\, .\\
\noalign{\vskip12pt}
{\rm (b)}\hskip5pt {\rm Re}(4-15\,i)=4\, ,\qquad {\rm Im}(4-15\,i)=-15\, .\\
\noalign{\vskip12pt}
{\rm (c)}\hskip5pt {\rm Re}(\cos\theta-i\,\sin\theta)=\cos\theta\, ,\qquad {\rm
Im}(\cos\theta-i\,\sin\theta)=-\sin\theta\, .\\
\noalign{\vskip12pt}
{\rm (d)}\hskip5pt i^2=-1.\quad  {\rm Re}(i^2)=-1\, ,\qquad {\rm Im}(i^2)=0\, .\\
\noalign{\vskip12pt}
{\rm (e)}\hskip5pt i\,(2-5\,i)=5+2\,i.\qquad  {\rm Re}(5+2\,i)=5\, ,\qquad {\rm Im}(5+2\,i)=2\, .\\
\noalign{\vskip12pt}
{\rm (f)}\hskip5pt (1+2\,i)(2-3\,i)=2-3\,i+4\,i+6=8+i\, .\qquad {\rm Re}(8+i)=8\, ,\qquad {\rm
Im}(8+i)=1\, .
\end{array}
$