Roots of Unity

**CropDuster** · January 22nd 2011, 12:35 PM

Hello, I have a basic (I think) question about Roots of Unity. First I feel like I should mention that I have only the most basic understanding of complex numbers. Here is what I have been asked to prove:

An $n$ th root of unity is a complex number $z$ such that $z^{n}=1$ . Prove that the $n$ th roots of unity form a cyclic subgroup of $\mathbb{C}^{\times}$ of order $n$ .

This is my first exposure to roots of unity, but after looking through a number theory book, I feel like the first sentence should include "...where $n\in\mathbb{Z}^{+}$ . Am I right about this?

If I am correct about this then how do I prove that each $n$ th root of unity has an inverse?

Then, how do I express the identity element of $\mathbb{C}^{\times}$ ?

I am feeling pretty lost here. Thanks in advance.

**Plato** · January 22nd 2011, 01:36 PM

The $n~~n^{th}$ roots of unity are $\xi _k = e^{i\frac{{2\pi k}}{n}} ,\;k = 0,1 \cdots ,n - 1$ .
Now $\xi _0 =1$ is the identity.
Also $\xi _1 = e^{i\frac{{2\pi}}{n}}$ is the generator.

**Ackbeet** · January 22nd 2011, 01:38 PM

Well, if you can assume that $\mathbb{C}^{\times}$ is a group, then you only have two things you have to prove: closure and inverses.

Yes, I would definitely say that $n\in\mathbb{Z}^{+}.$

For inverses, think about a root of unity, $z^{j},$ where $z^{n}=1.$ What do you have to multiply $z^{j}$ by to get $1?$

For closure, the product of two roots of unity must be a root of unity. So, take $z^{j}$ and $z^{k}.$ The product is $z^{j+k}.$ What happens when you raise this product to the $n$ th power?

**LoblawsLawBlog** · January 22nd 2011, 01:53 PM

The problem isn't just showing that each root has an inverse (all roots are nonzero and so have inverses in C), but that each inverse is also a root of unity. But that's really not too hard if you know enough about complex numbers. Specifically, I'm thinking that for any complex number $z$ with absolute value 1, $z\overline{z}=1$ , combined with the fact that $z^n-1=0$ and so $\overline{z}^n-1=$ ...

If this is more than you know, then it's not hard to list all nth roots of unity explicitly and then figure out the inverses that way. Also, if you know how the nth roots of unity are located in the complex plane and how multiplication of complex numbers adds angles and multiplies absolute values, then finding at least one generator shouldn't be too bad.

If you're not sure about the geometric view of complex numbers and their operations I'd suggest learning that because it makes problems like this so much easier to visualize.

**CropDuster** · January 22nd 2011, 03:19 PM

How is this? We are still in the first week of this class (abstract algebra) so I'm not sure how rigorous he expects us to be.

Let $U$ be the set of $n$ th roots of unity, with elements $a$ and $b$ . Then $a^{n}=1$ and $b^{n}=1$ . If we multiply both sides of $a^{n}=1$ on the right by $b^{n}$ we have

$a^{n}b^{n}=b^{n}$
$(ab)^{n}=b^{n}$

But since $b^{n}=1$ we have $(ab)^{n}=1$ and therefore $ab$ is an $n$ th root of unity. Hence, $ab\in U$ .

Since 1 is an identity element in $\mathbb{C}^{\times}$ and $1^{n}=1$ , the identity element in $\mathbb{C}^{\times}$ is also in $U$ .

Finally, let the inverse of $a^{n}$ be $a^{-n}$ . Since $a^{-n}=(a^{n})^{-1}=1^{-1}=1$ , we have that $a^{-n}=1$ . Hence, $a^{-n}\in U$ .

**Ackbeet** · January 22nd 2011, 04:34 PM

You need to show that $a^{-1}\in U,$ not $a^{-n}\in U.$

**CropDuster** · January 23rd 2011, 07:07 AM

Ah yes, well that is a glaring mistake! Thanks for your help. I've got a feeling that I'll have several more questions throughout the semester.

Finally, let the inverse of $a$ be $a^{-1}$ . Since $(a^{-1})^n=a^{-n}=(a^{n})^{-1}=1^{-1}=1$ , we have that $a^{-1}$ is an $n$ th root of unity. Hence, $a^{-1}\in U$ .

**Ackbeet** · January 24th 2011, 05:24 AM

Right. That looks better.

**CropDuster** · January 26th 2011, 06:41 AM

Hello again,

I believe that I have found a problem with my proof. While I have shown that a set $U$ consisting of the $n$ th roots of unity form a subgroup of $\mathbb{C}^{\times}$ , I have not shown that $U$ is cyclic, or that it is of order $n$ .

I am kinda stumped as to how I should go about showing this. While it seems obvious that my proof holds for any powers of $a$ , the issue seems that the order of $U$ could be infinite.

Thoughts?

**Ackbeet** · January 26th 2011, 06:47 AM

Take a closer look at Plato's post # 2. There's a generator there.

**LoblawsLawBlog** · January 26th 2011, 07:28 AM

As for possibly being infinite, there are several ways to show that it isn't. You can argue that there are only finitely many places the root can lie on the unit circle. For example, one 6th root of unity makes an angle of 360/6 degrees with the real axis. This will "go around" the circle once on its trip to 1. But a point can also go around twice. This would correspond to an angle of 360/3. Sooner or later you start to repeat points.

Also, you can show that an infinite cyclic group is isomorphic to $\langle\mathbb{Z}, +\rangle$ . In that group the equation $x^n=1$ would be written $nx=0$ using additive notation. This equation only has one solution in $\langle\mathbb{Z}, +\rangle$ , whereas the corresponding equation in the group of roots of unity has at least n.

Of course the easiest way is just to invoke the fact that an nth degree polynomial in any field can have at most n zeros. If you're allowed to assume this then it's the easiest way to go.

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