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More Roots of Complex Numbers - Problem 3
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I want to talk about a special problem that involves the roots of a complex number. Finding the nth roots of unity.

An nth root of unity is a complex number z satisfying the equation; z to the n equals 1. For example, the fourth roots of unity would be numbers satisfying z to the 4 equals 1. Let's find the cube roots of unity. These are the third roots of unity. These would satisfy z cubed equals one.

The first thing we have to do is express 1 in trigonometric form. 1 is a complex number, so you can do this. The modulus is of course going to be 1, and the argument is going to be 0; Cosine 0 plus i sine 0. What's the first root of unity going to be? Well I need to take the modulus of the original number and take its third root. Now the third root of 1 is 1. The modulus of all the roots is going to be one.

What's the argument of the first root? You take this number 0 and divide by 3. 0 divide by 3 is 0, so I get 0. Of course this is exactly the same number as I started with, so I get 1 and of course that makes sense. 1 cubed is 1. 1 should be one of the cube roots of unity.

Let's look for the second one. Remember there are going to be 3 of them. z sub 2 is going to be; same modulus, 1. What do we add to 0 to get the next root? Well we have to add 2 Pi divided by, whatever the index of the root is, 3, well 2 Pi over 3. I'm going to add 2 Pi over 3 to this and I get cosine of 2 Pi over 3 plus i sine 2 Pi over 3.

The cosine of 2 Pi over 3 is -1/2. Sine of 2 Pi over 3 is root 3 over 2. -1/2 plus root 3 over 2i. That's going to be another cube root of unity.

What about the third? 1, that's the modulus, cosine of; I add another 2 Pi over 3 to this. I get 4 Pi over 3 plus i sine 4 Pi over 3. Cosine of 4 Pi over 3, -1/2. Sine of 4 Pi over 3 is negative root 3 over 2. So I get negative one-half negative root 3 over 2 i.

The cube roots of unity are; 1, -1/2 plus root 3 over 2i and -1/2 minus root 3 over 2i. Any of these numbers if cubed gives 1.

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