Help with proof (Roots of unity question)

[Duckbeam]

1. The problem statement, all variables and given/kno(It n data

I need to prove that \prod_{k=1}^{n-1}2\sin\tfrac{k\pi}{n}=n
I just don't know where to start and what type of proof to do.

Is there any help to get? :D

Homework Equations

I know this has nothing to do with an equation.
But i have come up with the equation i have to prove after investigating lenths between roots x^n=1 where x=ℂ.

I wonder if I can somehow connect the factorization of x^n=1 to the proof.

The Attempt at a Solution

I have no idea what to do, this is why I am asking for help :D

 

[HallsofIvy]

One way of defining "sine" is to say that sin(t) is the y coordinate of the point on the circumference of the unit circle at distance t, around the circumference, from (1, 0). Here, t is k\pi/n so these points are equally spaced about the unit circle and, I am sure you understand, those points are the complex roots of the equation z^n= 1. Now, what can you say about the product of the imaginary parts of those roots?

(Note (x- a)(x- b)= x^2- (a+ b)x+ ab, (x- a)(x- b)(x- c)= x^3- (a+ b+ c)x^2+ (ab+ ac+ bc)x- abc, etc.)

 

[Duckbeam]

Here, t is k\pi/n so these points are equally spaced about the unit circle and, I am sure you understand, those points are the complex roots of the equation z^n= 1. Now, what can you say about the product of the imaginary parts of those roots?

What if k=1 --- this point isn't a root of z^n= 1? As far as i understand?

__

Now to your question I am also a little confused.

I take z^3=1 as an example. It has the roots e^{ik2\pi/3}. If I exclude the root at (1,0) I have two imaginary parts left: i\sin2\pi/3 and i\sin4\pi/3. So: \pm \sqrt{3}/2. The product of these being 3/4 --> not sure what you are getting at. But i might misunderstand it all.

 

[HallsofIvy]

Hmmm- you know I wondered briefly about this but did not follow up on it- what you are trying to prove is not true!

If n= 3, then k= 1 and 2 so that the product is 2sin(\pi/3)sin(2\pi/3)= 2(sqrt{3}/2)^2= 3/2, not 3!

 

[Duckbeam]

Hmmm- you know I wondered briefly about this but did not follow up on it- what you are trying to prove is not true!

If n= 3, then k= 1 and 2 so that the product is 2sin(\pi/3)sin(2\pi/3)= 2(sqrt{3}/2)^2= 3/2, not 3!

--> 2sin(\pi/3)2sin(2\pi/3)= 3

Did i write it wrong in my first post since you forgot the two in front of sin(2\pi/3)?

 

[Duckbeam]

I hope you didnt give up on me

 

[Duckbeam]

Anyone willing to help here?

 

[Dick]

It's a bit of a trick. x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1). You can also write x^n-1=(x-1)(x-x_1)(x-x_2)...(x-x_{n-1}) where the x_k are the other roots x^n-1 that aren't 1. So define f(x)=(x^{n-1}+x^{n-2}+...+x+1)=(x-x_1)(x-x_2)...(x-x_{n-1}). So f(1)=n and it's also equal to (1-x_1)(1-x_2)...(1-x_{n-1})=|1-x_1| |1-x_2| ... |1-x_{n-1}|. Now work out what those absolute values are.

 

[Duckbeam]

It's a bit of a trick. x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1). You can also write x^n-1=(x-1)(x-x_1)(x-x_2)...(x-x_{n-1}) where the x_k are the other roots x^n-1 that aren't 1. So define f(x)=(x^{n-1}+x^{n-2}+...+x+1)=(x-x_1)(x-x_2)...(x-x_{n-1}). So f(1)=n and it's also equal to (1-x_1)(1-x_2)...(1-x_{n-1})=|1-x_1| |1-x_2| ... |1-x_{n-1}|. Now work out what those absolute values are.

Yes i see somewhat of a connection here.

If i do it for n=3,4,5:

n=3 ---> (x-1) (x^2-x(2cos(2π/3)+1)
n=4 ---> (x-1)(x+1)(x^2-x(2cos(π/2)+1)
n=5---> (x-1)(x^2-x(2cos(2π/5)+1)(x^2-x(2cos(4π/5)+1

The connection is see is that when x=1 then if (x-1) is excluded it equals to n. I am not excactly sure why though.

 

[Dickfore]

Hint:
<br /> \prod_{k = 1}^{n - 1}{2 \, \sin \left( \frac{k \, \pi}{n} \right)} = \st{2 \, \Im \prod_{k = 1}^{n - 1}{\exp \left( i \, \frac{k \, \pi}{n} \right)}} = \ldots<br />
What is the sum in the exponential? What is \Im \exp \left( i \, \alpha \right), \ \alpha \in \mathbb{R}? Then, use the values of the sine.

EDIT:

The 2nd step does not follow from the first step. Namely: \mathrm{Im}{z_1} \, \mathrm{Im}{z_2} \neq \mathrm{Im}(z_1 \, z_2)

 

[Duckbeam]

Hint:
<br /> \prod_{k = 1}^{n}{2 \, \sin \left( \frac{k \, \pi}{n} \right)} = 2 \, \Im \prod_{k = 1}^{n}{\exp \left( i \, \frac{k \, \pi}{n} \right)} = 2 \, \Im \exp \left( i \, \frac{\pi}{n} \, \sum_{k = 1}^{n}{k} \right)<br />
What is the sum in the exponential? What is \Im \exp \left( i \, \alpha \right), \ \alpha \in \mathbb{R}? Then, use the values of the sine.

:( -- I am not completely familiar with all the notation here.

Especially the (exp) and the imaginary part. I am not sure how it all works together.
Any chance you can elaborate a little?

Also.. in the product it goes from k=1 to k=n-1 right? Thats not what you have written above -- is that on purpose or a typo?

 

[Duckbeam]

I really need help with this ASAP!

This proof is only PART of my huge assignment that is due in two days and i can't get it down. I really need some big time help here.

 

[Dick]

I really need help with this ASAP!

This proof is only PART of my huge assignment that is due in two days and i can't get it down. I really need some big time help here.

I'll do the case n=3. x^3-1=(x-1)(x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3}). Also x^3-1=(x-1)(x^2+x+1). Therefore (x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3})=(x^2+x+1). Put x=1 in that last expression and take the absolute value. Now try showing e.g. that | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3}).

 

[Duckbeam]

I'll do the case n=3. x^3-1=(x-1)(x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3}). Also x^3-1=(x-1)(x^2+x+1). Therefore (x-e^\frac{2 \pi i}{3})(x-e^\frac{4 \pi i}{3})=(x^2+x+1). Put x=1 in that last expression and take the absolute value. Now try showing e.g. that | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3}).

Yes yes i have done this!

When n=3 --> then x^2+x+1 = x^2-x(2cos(3π/2))+1
Pluggin in one --> 3 = 2-2cos(3π/2)
Using DA formula--> 3 = 2(1-(1-sin^2(π/3))

So 3 = 4sin^2 (π/3)

Now in my assignment i have already shown that | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3}) .. A quick way (for now) is to say that e^\frac{2 \pi i}{3} is the complex conjugate of e^\frac{4 \pi i}{3} hence they have the same distance..

so they are both 2sin(π/3) long.

But my whole dilemma in all this is that i have no understanding how this all fits together.
I think it is cool that when taking out (x-1) the remaining is equal to n when putting in x=1.. but i don't understand why.. would it work putting in any other root? And why do we take out (x-1) --

i need some kind of explanation so i can get understanding of the phenomenon! :D haha

 

[Dick]

Yes yes i have done this!

When n=3 --> then x^2+x+1 = x^2-x(2cos(3π/2))+1
Pluggin in one --> 3 = 2-2cos(3π/2)
Using DA formula--> 3 = 2(1-(1-sin^2(π/3))

So 3 = 4sin^2 (π/3)

Now in my assignment i have already shown that | 1 - e^\frac{2 \pi i}{3} | = 2 sin(\frac{\pi}{3}) .. A quick way (for now) is to say that e^\frac{2 \pi i}{3} is the complex conjugate of e^\frac{4 \pi i}{3} hence they have the same distance..

so they are both 2sin(π/3) long.

But my whole dilemma in all this is that i have no understanding how this all fits together.
I think it is cool that when taking out (x-1) the remaining is equal to n when putting in x=1.. but i don't understand why.. would it work putting in any other root? And why do we take out (x-1) --

i need some kind of explanation so i can get understanding of the phenomenon! :D haha

The whole thing works the same way for n not equal to 3. In general you will need to prove that | 1 - e^\frac{2 \pi i k}{n} | = 2 \sin \frac{k \pi}{n}. Those are the factors in your product. Or show | 1 - e^{i \theta} | = 2 \sin \frac{ \theta }{2} for 0 \le \theta \le 2 \pi.

 

[Duckbeam]

The whole thing works the same way for n not equal to 3. In general you will need to prove that | 1 - e^\frac{2 \pi i k}{n} | = 2 \sin \frac{k \pi}{n}. Those are the factors in your product.

I think i get it! I appreciate it! :D

 

[Duckbeam]

Man, i can't get it for general term n.. i don't know why

Any other hint that can help help me.

This is what i have found for n=3,4,5

n=3 : 4sin^2(π/3) ..
n=4 : 8sin^2(π/4)..
n=5 : 16sin^2(π/5)sin^2(2π/5)

I noticed this pattern:

2^n-1 sin^2(π/n) sin^2(2π/n)... sin^2(mπ/n) where m is the number of complex conjugatess..

 

[Dick]

Man, i can't get it for general term n.. i don't know why

Any other hint that can help help me.

This is what i have found for n=3,4,5

n=3 : 4sin^2(π/3) ..
n=4 : 8sin^2(π/4)..
n=5 : 16sin^2(π/5)sin^2(2π/5)

I noticed this pattern:

2^n-1 sin^2(π/n) sin^2(2π/n)... sin^2(mπ/n) where m is the number of complex conjugatess..

It looks like you are ignoring the hint I have you and trying to work it out case by case. That's a lot of work and you don't have to do it. The basic part of the hint was that (x^{n-1}+x^{n-2}+...+x+1)=(x-e^\frac{2 i \pi}{n})(x-e^\frac{i 4 \pi}{n})...(x-e^\frac{i 2 (n-1) \pi}{n}). Don't you believe it?

 

[Duckbeam]

It looks like you are ignoring the hint I have you and trying to work it out case by case. That's a lot of work and you don't have to do it. The basic part of the hint was that (x^{n-1}+x^{n-2}+...+x+1)=(x-e^\frac{i \pi}{n})(x-e^\frac{i 2 \pi}{n})...(x-e^\frac{i (n-1) \pi}{n}). Don't you believe it?

I do believe it! Because when x=1 the absolute values of the brackets on the RHS is basically the value of each distance and hence when multiplied together equals to n. But i still don't understand how to prove it.

i don't know if I am just stupid here.. but i can't see it at all.. :(btw, on the RHS i think you mean pi instead of x?

 

[Dick]

I do believe it! Because when x=1 the absolute values of the brackets on the RHS is basically the value of each distance and hence when multiplied together equals to n. But i still don't understand how to prove it.

i don't know if I am just stupid here.. but i can't see it at all.. :(

If you know the polynomial x^n-1 has n roots r_1, r_2, ... r_n, then you can factor the polynomial into x^n-1 = (x-r_1)(x-r_2)...(x-r_n). But you do know those n roots. They are 1, e^\frac{2 i \pi}{n}, e^\frac{4 i \pi}{n}, .... They are the nth roots of unity. BTW I missing a '2' in the roots in my previous post. I'll fix it.

But now there is another way to factor x^n-1 = (x-1) (x^{n-1}+x^{n-2}+...+x+1). Now just compare the two factorizations and 'cancel' the x-1 part.