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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.

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

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

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    $\begingroup$ cool animation (+1)! how did you make it? $\endgroup$ – J. W. Tanner Jun 14 at 3:35
  • $\begingroup$ I suggest you make it clear the circle has a radius of $1$, is centered on the origin, and is on the complex plane. $\endgroup$ – John Omielan Jun 14 at 3:36
  • $\begingroup$ Wait for a while, I am working on it. $\endgroup$ – nmasanta Jun 14 at 3:37
  • $\begingroup$ 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. $\endgroup$ – John Omielan Jun 14 at 3:40
  • $\begingroup$ @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 $1/n$. $\endgroup$ – John Omielan Jun 14 at 4:19
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An $n$th root of unity is a complex number $z$ which satisfies $$ z^n = 1. $$

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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.

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  • $\begingroup$ What do you mean $\pi$ is $\pi/2$?! Did you mean $-\pi$? $\endgroup$ – J. W. Tanner Jun 14 at 3:33
  • $\begingroup$ @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. $\endgroup$ – Eric Towers Jun 14 at 3:35
  • $\begingroup$ thanks for fixing $\pi$ is $\color{red}2\pi/2$ $\endgroup$ – J. W. Tanner Jun 14 at 3:42

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