<PRE><B>Question 453423</B>
(6+4i)(8-2i)

Ok, working with complex numbers is not too different than working with real numbers.  We just have to remember that a complex number has both a real component and an imaginary component and that i=sqrt(-1) and i^2=-1, etc.  Sooooooo
expand using FOIL (First, Outer, Inner, Last)
(6+4i)(8-2i)=48-12i+32i-8i^2 and this equals
48+20i-(8*(-1))=
48+20i+8=
56+20i---ans


hope this helps---ptaylor
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