Dividing Complex Numbers - Concept

Dividing by a complex number or a number involving i. So whenever we're dividing by a number that involves i, what we have to do is rationalize the denominator. Remember that i is equal to the square root of -1 and we're not allowed to have square roots in the denominator so we have to get rid of it. Okay. So we're going to go back to a problem that we already know how to do. 6 over root 8. So whenever we're dealing with a problem like this we have to rationalize the denominator. Get rid of that square root. So there's two ways of doing it. You could either multilply by root 8 over root 8 and get rid of that or what I tend to do is I like dealing with smaller numbers so if I can I try to simplify that denominator first.
I know that 8 is the same thing as 4 times 2. Let's do a different color so we can see it. So what this is actually really equal to is 6 over 2 root 2. So now instead of having them multiply by root 8, I still need to get rid of a radical but I can multiply by root 2 instead. So we multiply by root 2 and then [IB] to get to the square root and square the 2 in the top as well. Okay.
Before I multiply that through I can see that I can simplify this. We have 6 over 2. This is going to cancel leaving me with 3. Okay? So we now have 3 root 2 in the numerator and then we have the 2 is gone away. So we have root 2 over times root 2. Square roots. When you multiply them together they just cancel each other out leaving us with what's inside which is 2. So what we ended up with is 3 root 2 over 2. Okay?
So same exact idea when we are dealing with imaginary numbers, numbers involving i. So right here we have 5 over square root of 9. The first thing I want to do is to simplify that denominator radical, okay? This is square root of 9 is 3. So this is going to be 3i in the denominator. Okay? So rewriting this we have 5 over 3i. The 3 isn't presenting a problem, so we can leave it as this but what we really want to do is get rid of that i. Remember that i times i, i squared is -1. So if we multiply this by i ihn the denominator, we'll get i squared, -1. Our square root is gone. We have to multiply by 1, so we need an i in the top as well. Simplifying this out we got 5i in the numerator over 3i squared in the denominator. i squared, -1 so this just becomes -5i over 3 okay?
So just like we did with normal radicals, whenever we're dealing with the radical of a negative we still have to get rid of it. I find it best to simplify my numbers so I deal with smaller things. But the main problem is is to get rid of that square root in the denominator.