SOLUTION: 4+9i over 8-3i

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Question 44655: 4+9i over 8-3i
Answer by adamchapman(301) About Me  (Show Source):
You can put this solution on YOUR website!
So we want to know what %284%2B9i%29%2F%288-3i%29 is in its simplest form.
The problem is that we cannot just divide both terms on the top of the fraction above because the denominator is not just a simple number.
We need to eliminate the "imaginary" term, "i".
Try expanding %28a%2Bb%29%28a-b%29.
You will get a%5E2-b%5E2. Now if you multiply both the top and bottom of the complex fraction we started with, we will eliminate the imaginary tyerm in the denominator, and not alter the actual value of the fraction itself. The square of i is equal to negative one.
So multiply top and bottom by the "complex conjugate" of the denominator.
The complex conjugate of (a+bi) is (a-bi) (you just change the sign of the imaginary term to find the complex conjugate).
Here is the working:
%284%2B9i%29%2F%288-3i%29=%28%284%2B9i%29%288%2B3i%29%29%2F%28%288-3i%29%288%2B3i%29%29

%2832%2B84i%2B27i%5E2%29%2F%288-9i%5E2%29=%2832%2B84i-27%29%2F%288-9%29
%2832%2B84i-27%29%2F%288-9%29=%285%2B84i%29%2F-1
Finally:
%284%2B9i%29%2F%288-3i%29=-5-84i

I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam