# What is the meaning of “nth root of unity”? [closed]

I want to know the real meaning of nth root of unity. I have searched various books , websites and videos but couldn't find a satisfying answer. Every place where I tried to find my answer is just telling what is it's formula. Kindly help me out BTW this is not my homework.

## closed as off-topic by RRL, Cesareo, GNUSupporter 8964民主女神 地下教會, Xander Henderson, José Carlos SantosJun 14 at 18:09

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• its $n^{th}$ power is $1$ (it could be a complex number or in modular arithmetic, depending on the context) – J. W. Tanner Jun 14 at 2:51
• The word unity is a synonym of the number $1$. We want the $n$th roots of 1, i.e. solutions to $z^n=1$. – user10354138 Jun 14 at 3:06

Geometrically, the $$n^{th}$$ roots of unity correspond to $$n$$ points evenly dividing up a circle.

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Discussion:

Let $$x=1^{1/n}$$

$$\implies x=(\cos 0 +i \sin 0)^{1/n}=\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n},\qquad \text{where}\quad k= 0, 1, 2, . . ., n-1$$

Now, complex values can be graphed on the Cartesian coordinate system on $$x + iy \equiv (x,y)$$ (this is called the complex plane). Since we are mapping $$\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}$$ to $$x + iy$$, this gives us:

$$x =\cos \frac{2k\pi}{n}= \cos (2\pi\frac{k}{n})$$

$$y =\sin \frac{2k\pi}{n}= \sin (2\pi\frac{k}{n})$$

In Cartesian coordinate, the equation for a unit circle at $$(0,0)$$ is $$x^2+y^2=1$$, which is satisfied by our $$x\quad \text{and} \quad y$$. So we can say that each of the roots above maps to a point on the circumference of a unit circle.

So, all we have left to prove is that each of these $$n$$ points is equidistant from the adjacent points on the circle.

Clearly, we have points at based on the following $$n$$ values:

$$2\pi\frac{0}{n},\quad 2\pi\frac{1}{n},\quad 2\pi\frac{2}{n}, \quad. . . , \quad 2\pi\frac{n-1}{n}$$

Now consider a circle which has the radius $$r = 1$$.

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It is clear that plotting lines at $$2\pi\frac{0}{n},\quad 2\pi\frac{1}{n},\quad 2\pi\frac{2}{n}, \quad. . . , \quad 2\pi\frac{n-1}{n}$$ divides up the total circle ($$2π$$ radians) into $$n$$ equal portions.

Since $$\sin θ = \frac{y}{r} = y\quad \text{and}\quad \cos θ = \frac{x}{r} = x$$, it is clear that $$x=\cos (2\pi\frac{k}{n}) \quad \text{and}\quad y = \sin (2\pi\frac{k}{n})$$ are the places of intersection when the circle is evenly divided.

In other words, each $$x=\cos (2\pi\frac{k}{n}) \quad \text{and}\quad y = \sin (2\pi\frac{k}{n})$$ is a point at the place where the $$(\frac{k}{n})^{th}$$ part of the circle sweeps out against the circumference of the circle.

Since the length of the circumference is $$2\pi r^2 = 2(1)^2 π$$, this means that the length of each subtended arc is $$2π \frac{k}{n}$$.

This results in the patterns above depending upon the value of $$n$$.

Thanks to "Larry Freeman"

• cool animation (+1)! how did you make it? – J. W. Tanner Jun 14 at 3:35
• I suggest you make it clear the circle has a radius of $1$, is centered on the origin, and is on the complex plane. – John Omielan Jun 14 at 3:36
• Wait for a while, I am working on it. – nmasanta Jun 14 at 3:37
• No worries. One other thing you may wish to mention is that $(1,0)$ is always one of the points, with this then fixing the positions of the other $n-1$ points. Also, don't forget that to ping me with a reply, use "@" followed by my username. I only noticed you replied so quickly because I was still on this page. – John Omielan Jun 14 at 3:40
• @nmasanta Excellent, detailed answer (+1). However, you have a small typo in the line which starts with $\implies x=(\cos 0 +i \sin 0)^{1/4}=$, where the $1/4$ power should be $1/n$. – John Omielan Jun 14 at 4:19

An $$n$$th root of unity is a complex number $$z$$ which satisfies $$z^n = 1.$$

The phrase "$$n^\text{th}$$ roots of unity" is naturally placed in the context of complex numbers. One should set one's expectations: you have asked "what are [these numbers]?" If someone were to ask you "What are the even numbers?" what sort of answer could you give that is not a formula or a formula in disguise (e.g., the set of numbers $$2n$$ where $$n$$ is an integer)?

The square roots of unity are all the numbers whose square is $$1$$. There are two: $$\pm 1$$. Notice that their complex angles are evenly spaced around the circle, at angles $$0$$ and $$\pi$$ (which is $$2\pi/2$$, half a circle). Also, their magnitudes are all $$1$$.

The cube roots of unity are all the numbers whose cube is $$1$$: $$\dfrac{1}{2} + \mathrm{i}\dfrac{\sqrt{3}}{2}$$, $$\dfrac{1}{2} - \mathrm{i}\dfrac{\sqrt{3}}{2}$$, and $$1$$. Again, these have evenly spaced complex angles, $$0$$, $$2\pi/3$$, and $$4\pi/3$$ and their magnitude are all $$1$$.

In fact, the $$n^\text{th}$$ roots of unity all have magnitude $$1$$.

The $$4^\text{th}$$ roots of unity are the four numbers whose fourth powers are $$1$$. They are $$\pm 1$$ and $$\pm \mathrm{i}$$. Their complex angles are $$0$$, $$2\pi/4 = \pi/2$$, $$2\cdot 2\pi/4 = \pi$$, and $$3\cdot 2\pi/4 = 3\pi/2$$.

Perhaps you see the pattern. The $$n^\text{th}$$ roots of unity are the numbers whose $$n^\text{th}$$ power is $$1$$. There are $$n$$ of them. They have magnitude $$1$$ and their complex angles are multiples of $$2\pi/n$$. In polar form, these numbers have the form $$1 \cdot \mathrm{e}^{\mathrm{i} (k \cdot 2\pi /n)} \text{,}$$ for $$k = 0, 1, \dots, n-1$$, where the "$$1$$" is the magnitude, the "$$\mathrm{e}^{\mathrm{i} \dots}$$" encodes "with complex angle", and the complex angle is $$k 2\pi / n$$. This gives $$n$$ numbers. Let's look at one for $$n = 3$$ (using Euler's formula to convert from polar form to rectangular form): $$1 \cdot \mathrm{e}^{\mathrm{i} (1 \cdot 2\pi /3)} = \cos(1 \cdot 2\pi/3) + \mathrm{i} \sin(1 \cdot 2\pi/3) = \dfrac{1}{2} + \mathrm{i}\dfrac{\sqrt{3}}{2} \text{.}$$ And let's check that its cube really is unity (that is, $$1$$):\begin{align*} \left( 1 \cdot \mathrm{e}^{\mathrm{i} (1 \cdot 2\pi /3)} \right) ^ 3 &= 1^3 \cdot \left( \mathrm{e}^{\mathrm{i} (1 \cdot 2\pi /3)} \right) ^ 3 \\ &= 1 \cdot \mathrm{e}^{3 \mathrm{i} (1 \cdot 2\pi /3)} \\ &= 1 \cdot \mathrm{e}^{\mathrm{i} 2\pi} \\ &= 1 \cdot 1 \\ &= 1 \text{.} \end{align*}

What you generally find in references is that $$\xi_n = \mathrm{e}^{2\pi\mathrm{i}k/n}$$ for $$k = 0, 1, 2, \dots, n-1$$ is an $$n^\text{th}$$ root of unity and is a root of the polynomial $$x^n = 1$$. This says what we said above in many fewer words: an $$n^\text{th}$$ root of unity is a (complex) number whose $$n^\text{th}$$ power is unity ($$1$$), and those numbers have magnitude $$1$$ and proceed from $$1$$ anticlockwise by complex angle $$2\pi / n$$, meaning that their complex angles are evenly spaced.

• What do you mean $\pi$ is $\pi/2$?! Did you mean $-\pi$? – J. W. Tanner Jun 14 at 3:33
• @J.W.Tanner : Guessing about which context-free $\pi$ you are referencing, ..., I think it's fixed now. If that's the only surviving "$\pi$" versus "$2\pi$" transposition here, I'll be surprised. – Eric Towers Jun 14 at 3:35
• thanks for fixing $\pi$ is $\color{red}2\pi/2$ – J. W. Tanner Jun 14 at 3:42