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Dividing Complex Numbers - Problem 1
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Simplifying fractions. We know that we can’t have square roots in the denominator fractions so what we have to do is simplify this out. For this particular example we’re dealing with the square root of a negative number so we’re going to be dealing with i’s as well as a square root.

Whenever I see a problem like this the first thing I always want to do is simplify my square roots as much as possible. I like dealing with smaller numbers instead of bigger numbers. I look at this and I see that 4 goes into 20, square root of 4 is 2, so the numerator becomes 2 root 5. Looking at the denominator square root of 72. 72 can be divided up into 2 and 36, so this ends up being 6 root 2 and we also have the square root of -1 on the outside. Square root of -1 we know is i. So what we really have here is 2 root 5 times 6i root 2.

We not only need to get rid of the square root of 2 in the bottom but we also need to get rid of the i because that has a square root as well. In order to get rid of them we need to multiply by i root 2. i² becomes -1 root 2 times root 2 just becomes 2. We have to multiply by 1 in order to keep the problem balanced, so we need to multiply by i root 2 in the numerator as well. I’m not multiplying by 6 because that 6 isn’t presenting any type of problem, it’s just a number sitting there, 6 can be in the denominator just fine.

We now simplify everything out. We have 2 in the numerator, we have i in the numerator and our root 5 times root 2 is going to become root 10. The 6 stays there, i times i becomes i² which we know to be -1 and root 2 times root 2 becomes 2.

Simplifying this out, we have a 2 at the top and the bottom, those can cancel, 6 and the -1 we can combine so we end up with i root 10 all over -6.

By rewriting both of our square roots, multiplying till you get rid of our fraction, our square root, and our i in the bottom we were able to rationalize our denominator.

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