SOLUTION: 1. (3+2i)squared(3-2i)squared (3+2i)(3+2i)(3-2i)(3-2i) 9+6i+6i+4isquared 9-6i-6i+4isquared 9+12i+4(-1) 9-12i+4(-1) (5+12i)(5-12i)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 1. (3+2i)squared(3-2i)squared (3+2i)(3+2i)(3-2i)(3-2i) 9+6i+6i+4isquared 9-6i-6i+4isquared 9+12i+4(-1) 9-12i+4(-1) (5+12i)(5-12i)       Log On


   

Question 133210: 1. (3+2i)squared(3-2i)squared
(3+2i)(3+2i)(3-2i)(3-2i)
9+6i+6i+4isquared 9-6i-6i+4isquared
9+12i+4(-1) 9-12i+4(-1)
(5+12i)(5-12i)
=169
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
You are absolutely correct. There was, however, a slightly more elegant and certainly simpler way to do this one.

%283%2B2i%29%2Agreen%28%283%2B2i%29%29%2Ared%28%283-2i%29%29%283-2i%29

First, re-arrange the factors:
%283%2B2i%29%2Ared%28%283-2i%29%29%2Agreen%28%283%2B2i%29%29%283-2i%29

Now you have two sets of conjugates to multiply:
%283%2B2i%29%283-2i%29=9%2B4=13, so:

%283%2B2i%29%2Ared%28%283-2i%29%29%2Agreen%28%283%2B2i%29%29%283-2i%29=13%2A13=169

Like my Daddy always used to say, "There's more'n one way to skin a cat." Keep an eye out for these sorts of things. They will save you lots of time and frustration, particularly at test time.