what is the product of the complex numbers below (3-9i)(6-i)

Answer 1

Hey Demitri , great question!

Take the 3 at the beginning and times it into both the six and the negative "i" . You will get 18-3i. Then, do the same with the negative nine "i" to get -54i plus nine "i" SQUARED(because -9i times negative i=positive 9 i "squared") .Anytime you get an "i SQUARED" you change it up to (-1), so "i" squared, every time is always equal to negative one. Anytime you get "i squared", it is always equal to negative one.

Take 18-3i -54i+ 9i squared and put the -3i -54i together to make -57i (like terms) and then since the "i" squared = (-1), make "9 i squared" into 9 times (negative one) because that is what i squared is-negative one. So you got 18-57i-9, which u put the 18-9 together into just 9, so you have 9-57i. You got this!

Answer 2

Using FOIL we get 3x6=18 3 x -i= -3i -9i x6 = -54i. -9i x -i= +9i^=-9

Adding up the 4 terms we get 9-57i