What Are Roots of Unity?

Finding the $n$th roots of unity

Great! So we now know what a $n$th root of unity is.

How would we go about finding the $n$th roots of unity?

Lets use De Moivre's Formula!

$$\begin{aligned} x^n&=1=1(\cos 0+i\sin 0) \\x&=[1(\cos 0+i\sin 0)]^{-n} \\&=1(\cos -n0+i\sin -n0) \\&=1 \end{aligned}$$

This is expected, since $1^n=1$ for all $n$.

But according to the Fundamental Theorem of Algebra, there are $n$ solutions to $x^n=1$, so how would we find the other $n-1$ solutions?

Remember that $0$ rad is equal to $2k\pi$ rad for all integers $k$.

The rest of the solutions can be found when we replace the angle with $2k\pi$

Hence all solutions to $x^n=1$ will be:

$$\begin{aligned} x&=[1(\cos 2k\pi+i\sin 2k\pi)]^{-n} \\&=\left(\cos \frac{2k\pi}n+i\sin \frac{2k\pi}n\right) \end{aligned}$$

However, when $k=n$, the solution will be $(\cos 2\pi+i\sin 2\pi)$. Which is basically $1$.

Hence, for all $k\geq n$, the roots will repeat.

So therefore, the $n$th roots of unity are:

$$\left(\cos \frac{2k\pi}n+i\sin \frac{2k\pi}n\right)$$

Where $k\in\mathbb{Z}, 0 \leq k \leq n-1$

Roots of Unity

Here, we focus on the roots of a specific form of a polynomial, namely $x^n-1=0$. The roots of this polynomial form are called the Roots of Unity. Many interesting things can be found out about the roots of unity, and we will find a generalization for them in terms of n as well.

Let's begin with $x^3-1=0.$ We can quickly factor this with common factoring techniques to achieve $x^3-1=(x-1)(x^2+x+1)=0.$ Thence, the three roots are:

$1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}.$ In exponential notation, they are $e^0$, $e^{2\pi i/3}$, and $e^{4\pi i/3}$

We now plot them on the unit circle.

You get an equilateral triangle.

Interesting! Let's continue onto $x^4-1=0$. We can quickly factor and notice that $x^4-1=(x^2-1)(x^2+1)=0$. So, the four roots are:

1,i,-1,-i. In exponential notation, they are $e^0, e^{\pi/2}, e^{\pi i}, e^{3\pi i/2}$

Again, we plot them on the unit circle.

You get a square.

They can be proved, but now we list our generalizations.

For any given polynomial of the form $x^n-1=0$, The complex number $e^0, e^{2\pi i/n}, e^{4\pi i/n}, e^{6\pi i/n}, ... , e^{2(n-1)\pi i/n}$.

When graphed on the unit circle, these roots of unity will always, when connected, form a regular n-gon.

Lastly we explore primitive roots of unity.

An $nth$ root of unity, $z$, is called a primitive nth root of unity if $z^n=1$ and $z^k$ does not equal $1$ for $k=1,2,3,...n-1.$ As expected, we can make a generalization for these special roots of unity.

If $gcd(k,n)=1$, then $\omega=e^{2\pi ik/n}$ is a primitive nth root of unity, and $1, \omega, \omega^2,...\omega^{n-1}$ are all $nth$ roots of unity.

Roots of unity are interesting because they are both algebraic and trigonometric: they have a nice representation as the roots of a polynomial ($x^n-1$) and as complex numbers ($\text{cis} \frac{2\pi k}{n}$). We can often combine these to prove some surprising results; the following problem is an example of one.

Show that

$$\cos(0) + \cos(\pi/11) + \cos(2\pi/11) + \ldots + \cos(21\pi/11) = 0.$$

The key idea is to bring in complex numbers. This might seem odd - but as we'll see, this simplifies the problem greatly.

Each cosine term is a real part of a corresponding root of unity (specifically, here we have $22$nd roots of unity, since we have $2\pi\cdot0/22,$ $2\pi\cdot1/22,$ $2\pi\cdot2/22, \ldots$). Therefore, the entire equation is equivalent to

$$\text{Re}[\text{cis}(0) + \text{cis}(\pi/11) + \text{cis}(2\pi/11) + \ldots + \text{cis}(21\pi/11)] = 0.$$

Recall that these $22$ complex numbers are collectively all of the roots of the equation

$$x^{22} - 1 = 0.$$

We are looking at the sum of the roots of a polynomial - so Vieta's formulas is a natural fit. The sum is the negative of the $x^{21}$ coefficient, divided by the $x^{22}$ coefficient...

Yes, the sum of the roots of unity is $0$ by Vieta's formulas! Therefore, the sum of the cosines we were given, which is the sum of the real part of our roots of unity, must also equal $0.$ We're done! $\square$

In general, the sum of the $n$th roots of unity always equals $0$ - see why?

(Video) Example of Roots of Unity

by PatrickJMT

In this video by PatrickJMT, you will learn about an example of the Roots of Unity.

Roots of Unity

The roots of unity are special complex numbers that, when raised to a certain power, equal $1$.

For example, the square roots of unity are the numbers that satisfy $z^2=1$. These are just $1$ and $-1$ because both numbers squared give $1$.

In general, the n^{th} roots of unity are the solutions to the equation: $$z^n = 1$$

This means that we are looking for $n$ complex numbers that, when multiplied by themselves $n$ times, give $1$.

One of these roots is always $z=1$, but the rest of the complex numbers that form a regular polygon on the unit circle in the complex plane. That's because all roots of unity have a magnitude, $r$, of $1$ and are evenly spaced around the circle.

We can write them using polar form: $$z_k = \text{cis}(\frac{2\pi k}{n})$$

, where $k=0,1,2,...,n-1$

These are called the "n-th roots of unity" because each one is a "unity" (i.e., has a magnitude of $1$), and together, they complete a full circle.