How do you multiply #(7-3i)(7+3i)#?
1 Answer
Jul 26, 2016
Explanation:
Note that for any
#a^2-b^2 = (a-b)(a+b)#
So we find:
#(7-3i)(7+3i) = 7^2-(3i)^2 = 49-9i^2 = 49+9 = 58#
Note that for any
#a^2-b^2 = (a-b)(a+b)#
So we find:
#(7-3i)(7+3i) = 7^2-(3i)^2 = 49-9i^2 = 49+9 = 58#