How do you evaluate #(5- 8i ) ( 5- 5i )#?

1 Answer
Nov 6, 2017

#-15 - 65i#

Explanation:

This can be done no differently than a regular polynomial multiplication problem. In the case of a binomial times a binomial (such as this problem "looks like") you can use the FOIL process (aka distributive operation) to multiply the product out, after which point you can use the properties of the imaginary unit #i# to simplify:

#(5-8i)(5-5i) = underbrace(5(5))_ "first" + underbrace(5(-5i))_ "outer" + underbrace((-8i)*5)_ "inner" + underbrace((-8i)*(-5i))_ "last"#

# = 25 - 25i - 40i + 40i^2 = 25 -65i + 40i^2#

Since #i^2 = -1#:

# = 25 - 65i + 40(-1) = 25 - 65i - 40 = -15 - 65i#