SOLUTION: multiply the complex numbers (3+2i)(3-2i) I think it is done like this. 9-6i+6i-4i^2 cancel out the 6's 9-4i^2 multiply the -4x-1 9+4 13+0i

Question 129163: multiply the complex numbers
(3+2i)(3-2i)
I think it is done like this.
9-6i+6i-4i^2
cancel out the 6's
9-4i^2
multiply the -4x-1
9+4
13+0i
Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
multiply the complex numbers
(3+2i)(3-2i)
I think it is done like this.
9-6i+6i-4i^2
Simplify to get:
9-4i^2
multiply -4(-1)
9+4
13+0i
-----------
You are correct.
Cheers,
Stan H.

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
13 is the answer. 0i=0
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