SOLUTION: Please help me with this problem: (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 a

Question 518014: Please help me with this problem:
(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?
Thanks for your help. It is much appreciated.
Found 2 solutions by MathLover1, Alan3354:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
......both sides multiply by

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
(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?
-------
-4*-1 = +4
------------
[(4i)/(3+i)] * (3-i)/(3-i) = (12i-4i^2)/(9-i^2) right so far
= (4 + 12i)/10
= (2 + 6i)/5

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