a)
The formula for multiplying complex numbers is
\\[\\begin{eqnarray*}\\simplify[]{Re((a + ib)(c + id))} &=& ac -bd \\\\ \\simplify[]{Im((a + ib)(c + id))} &=& ad +bc \\end{eqnarray*} \\]
So we have:
\\[\\begin{eqnarray*}\\simplify[]{Re({a}*{b})} &=& \\simplify[]{{Re(a)}*{Re(b)} - {Im( a)}*{Im(b)} = {Re(a*b)}}\\\\ \\simplify[]{Im({a}*{b})} &=& \\simplify[]{{Re(a)}*{Im(b)} + {Im( a)}*{Re(b)} = {Im(a*b)}} \\end{eqnarray*} \\]
Hence the solution is :
\\[(\\simplify[std]{{a}})(\\simplify[std]{{b}})=\\var{a*b}\\]
b)
This is calculated in a similar way once the expression is written as:
\n$(\\simplify[std]{{a1}})^2= (\\simplify[std]{{a1}}) (\\simplify[std]{{a1}})$ then we find:
\n\\[\\begin{eqnarray*}(\\simplify[std]{{a1}})^2&=& (\\simplify[std]{{a1}}) (\\simplify[std]{{a1}})\\\\ &=& \\simplify[]{({Re(a1)}*{Re(a1)} - {Im(a1)}*{Im(a1)})+ ({Re(a1)}*{Im(a1)} + {Im(a1)}*{Re(a1)})i}\\\\ &=& \\simplify[std]{{a1^2}} \\end{eqnarray*} \\]
c)
We know that $i^2=-1$ which gives $i^3=i^2i=-i$.
Hence:
\\[ \\begin{eqnarray*} \\simplify[std,!otherNumbers]{{a3} + {b3} * i + {c3} * i ^ 2 + {d3} * i ^ 3}&=&\\simplify[std]{{a3} + {b3} * i -{c3} -({d3} * i)}\\\\ &=&\\simplify[std]{ {a3} -{c3} + ({b3} -{d3}) * i}\\\\ &=&\\simplify[std]{{a3 -c3} + {b3 -d3} * i} \\end{eqnarray*} \\]
d)
This can be calculated by using the formula twice, firstly to multiply out the first two sets of parentheses,
and then to multiply the result of that calculation by the third set of parentheses.
So we obtain:
\\[ \\begin{eqnarray*} (\\var{z1})(\\var{z2})(\\var{z3})&=&((\\var{z1})(\\var{z2}))(\\var{z3})\\\\ &=&(\\var{z1*z2})(\\var{z3})\\\\ &=&\\var{z1*z2*z3} \\end{eqnarray*} \\]
$(\\simplify[std]{{a}})(\\simplify[std]{{b}})\\;=\\;$[[0]].
\n\n
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\n \t\t \t\t16/07/2012:
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