Saturday, October 22, 2011

How many cube roots of one are there?

When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.

The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root of negative one.

What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.

What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.

Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.

If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just looking at the other root.

Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How about we cube it.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))

First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)

We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).

(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)

What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean? The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!

1 - i/4√(3) + i/4√(3)
1 + 0

So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No! Mathematics is never random!

If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get 90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.

What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth roots, and the 60 degree marks give you the sixth roots.

If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know.

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