Multiplying Complex Numbers - Problem 1 3,514 views
Multiplying complex or imaginary numbers. Whenever we multiply complex and imaginary numbers they behave a lot like anything else that we multiply.
However what we have to do when we have to when we’re dealing with the square root of negatives is we’re used to being able to combine square roots of square roots. We have the square root of 3 times the square root of 5, we can rewrite this as the square root of 15. It’s a little bit more complicated when we’re dealing with the square root of negative numbers. What we have to do first is to rewrite them as i’s. We are going to have to simplify these up before we even combine them.
So what we can do is we can break this down and so this is going to be equal to the square root of 8 and the square root of -1 and we also have the square root of 50 times the square root of -1. Now we have a couple of approaches on how we could solve this out. What we could do is we could simplify the square root of 8 independently and the square root of 50 independently and then multiply those together or we could do what we did right over here and combine those two square roots but we have to deal with the -1's separately. That’s what I’m going to do, either one is going to be perfectly fine.
What that ends up giving us is the square root of 8 and the square root of 50 as one unit and we have square root of -1, square root of -1 as another. We’re going to have to rewrite these square roots of -1's as i’s, so what we really have over here is i times i which is the same thing as i². Square root of 8 times 50 is the same thing as the square root of 400. Square root of 400 is 20. So what we have here is 20 times i². i² if you remember is the same thing as -1, so we can substitute that in here, this is -1. We really have 20 times -1 which is -20.
If we would have multiplied these through together beforehand, our negatives would have cancelled and just given us the answer 20. We always, always, always have to rewrite in terms of i’s when we’re dealing with the square root of negative number, other than that it’s the same exact thing as we already know.