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

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
%284i%29%2F%283%2Bi%29......both sides multiply by 3-i

%28%284i%29%283-i%29%29%2F%28%283%2Bi%29%283-i%29%29

%2812i-4i%5E2%29%2F%289-i%5E2%29

%2812i-4i%5E2%29%2F%289-%28-1%29%29

%2812i-4i%5E2%29%2F%289%2B1%29

%2812i-4i%5E2%29%2F10

%28cross%2812%296i-cross%284%292%28-1%29%29%2Fcross%2810%295

%286i%2B2%29%2F5

2%2F5%2B6i%2F5


Answer by Alan3354(69443) About Me  (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?
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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