Video: Summing Powers of Primitive Roots of Unity

If πœ” is a primitive 6th root of unity, which of the following expressions is equivalent to πœ” + πœ”Β² + πœ”Β³? [A] πœ”β΄ + πœ”β΅ + πœ”βΆ [B] 1 βˆ’ πœ”β΄ βˆ’ πœ”β΅ [C] 1 [D] βˆ’(1 + πœ”β΄ + πœ”β΅) [E] (1/2)(πœ”Β² + πœ”β΄ + πœ”βΆ)

06:59

Video Transcript

If πœ” is a primitive sixth root of unity, which of the following expressions is equivalent to πœ” plus πœ” squared plus πœ” cubed? Option (A) πœ” to the fourth power plus πœ” to the fifth power plus πœ” to the sixth power. Option (B) one minus πœ” to the fourth power minus πœ” to the fifth power. Option (C) one. Option (D) negative one times one plus πœ” to the fourth power plus πœ” to the fifth power. Or option (E) one-half times πœ” squared plus πœ” to the fourth power plus πœ” to the sixth power.

In this question, we’re told that πœ” is a primitive sixth root of unity. And we need to use this information to determine an equivalent expression to πœ” plus πœ” squared plus πœ” cubed. Since πœ” is a primitive sixth root of unity and the expression we’re given involves summing powers of πœ”, we can start by recalling a very useful result about the sums of powers of roots of unity. We recall for any integer 𝑛 greater than one and if 𝑧 is a primitive 𝑛th root of unity, then one plus 𝑧 plus 𝑧 squared β€” and we sum all the way up to 𝑧 to the power of 𝑛 minus one β€” will be equal to zero. We’re going to want to use this fact to find something equivalent to the expression given to us in the question.

First, we’re told that πœ” is a primitive sixth root of unity, so we’ll set our value of 𝑛 equal to six. And then using πœ” as our primitive root of unity, we get one plus πœ” plus πœ” squared plus πœ” cubed plus πœ” to the fourth power plus πœ” to the power of five is equal to zero. And of course, we want to use this to find an expression for πœ” plus πœ” squared plus πœ” cubed. And we can see that this already appears inside of our expression. So we’ll rearrange our equation to make this the subject. We’ll do this by subtracting one from both sides of our equation. We’ll subtract πœ” to the fourth power from both sides of our equation. And we’ll subtract πœ” to the fifth power from both sides of our equation.

Doing this, we get that πœ” plus πœ” squared plus πœ” cubed is equal to negative one minus πœ” to the fourth power minus πœ” to the fifth power. And now we can simplify this even further. We can notice all three terms on the right-hand side of our equation share a factor of negative one. So we’ll take out the shared factor of negative one. This gives us negative one times one plus πœ” to the fourth power plus πœ” to the fifth power. And then we can see this is exactly the same as the answer given in option (D). So we could just stop here. However, there is one thing worth pointing out.

We might be wondering why this identity is true in the first place. And one way to see that this is true is to look at this geometrically. We’re going to plot this onto an Argand diagram. First, we’re going to want to use another fact about roots of unity. If 𝑧 is an 𝑛th root of unity, then the modulus of 𝑧 is going to be equal to one. And in fact, we can directly show this is true. First, if 𝑧 is an 𝑛th root of unity, then 𝑧 to the 𝑛th power will be equal to one. We’re then going to take the modulus of both sides of this equation. This gives us the modulus of 𝑧 to the 𝑛th power will be equal to the modulus of one. And of course, the modulus of one is just equal to one. And the modulus of 𝑧 to the 𝑛th power is equal to the modulus of 𝑧 all raised to the 𝑛th power.

Now we need to use the fact that the modulus of any number is always nonnegative. This is because the modulus represents the distance of our point from the origin on our Argand diagram, and distances can never be negative. So we have a nonnegative number raised to the 𝑛th power is equal to one. This means that the modulus of 𝑧 must be equal to one. So let’s think about what we’ve shown. We’ve shown for any 𝑛th root of unity, the modulus of this number has to be equal to one. And if their modules is equal to one, their distance from the origin on our Argand diagram is equal to one. In other words, all 𝑛th roots of unity lie on the unit circle centered at the origin on our Argand diagram.

And we’re going to need to use one more piece of information. We know how to generate all of the 𝑛th roots of unity. In particular, we can write them in polar form. We’ll write them in the form the cos of two πœ‹π‘˜ over 𝑛 plus 𝑖 times the sin of two πœ‹π‘˜ over 𝑛, where π‘˜ is an integer between zero and 𝑛 minus one. So we know all of our roots of unity lie on the unit circle. And we know one more thing; we know all of their arguments. Their arguments are all in the form two πœ‹π‘˜ over 𝑛, where π‘˜ is an integer between zero and 𝑛 minus one. We can then plot these onto our diagram. We’ll use the value of 𝑛 equal to six. When 𝑛 is equal to six, our arguments are going to be multiples of two πœ‹ divided by six, which we can simplify to give us πœ‹ by three.

So in this case, we’re going to get six points on our unit circle spaced evenly because their arguments all differ by πœ‹ by three. We’ll get a sketch which looks something like this. At this point, there’s several different choices for our primitive roots of unity. We’ll choose the primitive roots of unity to be when π‘˜ is equal to one. However, it wouldn’t actually matter which of our primitive roots of unity we chose, we would get the same answer. The only fact we’re going to use is that the primitive roots of unity can be used to generate all of the other roots of unity.

Now we want to add together all of these roots of unity on our Argand diagram. Now, we could do this directly from our formula for our roots of unity, and this would work. However, we can also do this geometrically. We recall we can add together points on our Argand diagram by instead considering vectors on our Argand diagram. For example, by adding together the two vectors shown on our Argand diagram, we would find the value of one plus πœ”. However, there’s a trick we can use to make these calculations easier. We’ll start by moving our vector from the origin to πœ” to instead point from πœ” to the fifth power to one.

Next, we’ll do something very similar to our vector pointing from the origin to one. We’re going to move it so it points from πœ” to the fourth power to πœ” to the fifth power. This gives us a different way of looking at one plus πœ”. It’s exactly the same as the vector which goes from πœ” to the fourth power to one. But remember, we want to find something equivalent to summing all of these values together. So we’re going to need to add on more vectors to our diagram. Let’s start by adding the vector from the origin to πœ” squared. Once again, we can simplify this by moving our vector. We’ll move this to point from one to πœ”. Next, we want to include πœ” cubed in our sum, so we need to add the following vector. And one way of doing this is to move our vector to point from πœ” to πœ” squared.

And now we’re starting to see a pattern. Doing exactly the same thing we did before, we can add the final two roots of unity into our diagram. And then after moving our vectors, we would end up with a diagram which looks like the following. The vectors end up forming a perfect hexagon inside of our unit circle. And in fact, we can directly show this is true just from the fact that their arguments differ by πœ‹ by three, and they lie on the unit circle. Now we can ask the question, what happens when we add all of these roots of unity together? We know we can do this by just adding our vectors together. Well, if we start at one and then we add all of our vectors together, we see we just end up at one. And if we start and end at the same point, then the horizontal and vertical components of our vector must be equal to zero.

In other words, because this vector is equal to zero, the sum of all of these roots of unity must also be equal to zero. This gives us a geometric reason why the sum of all of these roots of unity is equal to zero. And we were able to use this result to show if πœ” is a primitive sixth root of unity, then πœ” plus πœ” squared plus πœ” cubed will be equal to negative one times one plus πœ” to the fourth power plus πœ” to the fifth power, which was given as option (D).

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.