Question 518014
(4i)/(3+i) 
So far, I've got: 
[(4i)/(3+i)] * (3-i) = (12i-4i^2)/(9-i^2) 
I know that having i^2 = -1 but I was wondering if since it's already -4i^2 does that make it +4 or does it stay -4? 
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-4*-1 = +4
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[(4i)/(3+i)] * (3-i)/(3-i) = (12i-4i^2)/(9-i^2) right so far
= (4 + 12i)/10
= (2 + 6i)/5