<PRE><B>Question 527579</B>
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Use FOIL just like multiplying any other pair of binomials.  Just remember that *[tex \Large i^2\ =\ -1]


In general:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ +\ bi)(c\ +\ di)\ =\ ac\ +\ (ad\ +\ bc)i\ +\ (bd)i^2\ =\ ac\ +\ (ad\ +\ bc)i\ -\ bd\ =\ (ac\ -\ bd)\ +\ (ad\ +\ bc)i] 

John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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