Wolfram's MathWorld web site has a wonderful graphic that demonstrates this:

Here's the reason that it's true:

Theorem: The N-th Roots of Unity correspond to n points evenly dividing up a circle

Proof:

(1) The number of radians in a circle is 2π [See here for review of radians, if needed]

(2) The equation for roots of unity is (see Corollary 1.1, here):

where 0 ≤ k ≤ n-1.

(3) So each root of unity is:

cos[ (2kπ)/n] + isin[(2kπ)/n]

where 0 ≤ k ≤ n-1

(4) Which is the same as:

cos[(k/n)2π] + isin[(k/n)2π]

where 0 ≤ k ≤ n -1

(5) Now, complex values can be graphed on the Cartesian coordinate system on x + iy [this is called the complex plane].

(6) Each of the roots above maps to a point on the circumference of a unit circle since:

(a) The equation for a unit circle at (0,0) is [See Theorem 1, here]:

x

^{2}+ y

^{2}= 1

(b) Since we are mapping cos[(k/n)2π] + isin[(k/n)2π] to x + iy, this gives us:

x = cos[(k/n)2π]

y = sin[(k/n)2π]

(c) From a previous result (see Corollary 2, here), we have:

cos

^{2}[(k/n)2π] + sin

^{2}[(k/n)2π] = 1

(d) So, applying step #6b above gives us:

x

^{2}+ y

^{2}= 1

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

(8) Clearly, we have points at based on the following n values:

(0/n)2π, (1/n)2π, (2/n)2π, ..., [(n-1)/n]2π

where

y = cos[(k/n)2π]

x = sin[(k/n)2π]

(9) Now, if we use a unit circle, it is clear that each of these points is equidistant from each other since:

Consider a circle which has the radius r = 1.

It is clear that plotting lines at (0/n)2π, (1/n)2π. ... ([n-1]/n)2π divides up the total circle (2π radians) into n equal portions.

Using a unit circle, it is clear that x=[cos(k/n)2π] and y = [sin(k/n)2π] are the places of intersection when the circle is evenly divided since:

sin θ = y/r = y

cos θ = x/r = x

In other words, each y=cos[(k/n)2π] and x=sin[(k/n)2π] is a point at the place where the (k/n)th part of the circle sweeps out against the circumference of the circle.

Since the length of the circumference is πD=(2r)πr = 2(1)π, [see Corollary 1, here], this means that the length of each subtended arc is (k/n)*2π.

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

QED

For example, if we graph the third roots of unity. Then, we have:

lines sweeping out at:

0, (1/3)2π, and (2/3)2π.

which results in the following graph:

References

- Rowland, Todd and Weisstein, Eric W. "Root of Unity." From
*MathWorld*--A Wolfram Web Resource. http://mathworld.wolfram.com/RootofUnity.html - "Complex Plane", Wikipedia.com
- "Roots of Unity", Wikipedia.com

## 2 comments :

Wow! This is brilliant.... I just don't have words for it Larry. You are doing a wonderful job here. How come you manage to produce so many cool proofs man! I doubt you are a SW Engg. or Mathematician. Actually I stumbled upon your blog on search of a good explanation and proof for binomial theorem! But I was surprised to find everything in here, including math refresher. Thanx a lot man. Thats really a heck of a job you are doing. I myself is SW Engg. but never had a edge on math through out my life, I decided recently to give it another try so that I start loving it and and may be can write proofs like you :-). Do you have any suggestions?

Hi Nataraj,

Thanks very much for your comments! :-)

I am also a Software Engineer by profession.

Based on your write up, I don't really have any suggestions. It sounds like you are on your way.

Cheers,

-Larry

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