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MATH HELP NEEDED: Dividing imaginary numbers

dude8604

Platinum Member
I have a math assignment, and part of it has to do with dividing imaginary numbers. The textbook doesn't explain it very well, and I lost my math notes about it. Here is an example of one of the problems:

(12+8i) / (3+i)

Also, please explain how to do it, because I have to do a few more problems like it. Thanks!😎
 
The key is:

Rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator.

In other words, for the example given:

(12+8i) / (3+i) = [(12+8i)(3-i)]/[(3+i)(3-i)]

= (36 - 12i + 24i - 8i^2)/(9 - 3i + 3i - i^2)

which simplifies to

(44 + 12i)/(10)

which is

(22/5) + (6/5)i

in a+bi form.


edit: it's clearer (or at least to me) if you rewrite everything so that the eq's look like:

(12+8i)
(3+i)

etc

edit: aw h3ll

(12+8i) = (12+8i)(3-i)
(3+i)     (3+i)(3-i)

= (36 - 12i + 24i - 8i^2)
(9 - 3i + 3i - i^2)

which simplifies to

(44 + 12i)
(10)

which is

(22/5) + (6/5)i

in a+bi form.
 
Matt made a mistake in the first line of his example.
(12+8i) / (3+i) = [(12+8i)(3-i)]/[(3-i)(3-i)]
Should be
(12+8i) = (12+8i)(3-i)
(3+i) (3+i)(3-i)

Also the reasoning behind using the conjugate: to keep the fraction equal you must multiply both top and bottom by the same thing. In this case, that would by the complex conjugate. Why does this work? Well, by taking the original number and changing the sign, you are making it so that when the two are multiplied together, the middle terms of the FOIL cancel each other. This eliminates the i's. The beginning term is two real numbers multiplied together which results in a real number, and the end is i x -i = 1.
 


<< Thank you sooo much to ThisIsMatt and ddjkdg!😀 I'm giving you 10's (it's all I can do). >>

Whaaaaaaaaaat!?!? Raise my rating?!?!?! 😛

I'm pretty sure my solution is error free now 😛
 
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