#3**+10 **

They are the solutions to the 12th order polynomial x^{12} = 1

A 12th order polynomial has exactly 12 roots.

They can be pictured as equally spaced points on the unit circle in the complex plane.

Each root is separated from the next by an angle of 360/12 = 30 degrees.

Starting with the obvious root x = 1 + 0i on the x-axis, the next one (in a clockwise direction) has coordinates x = cos(30) and y = sin(30) (because the radius of the unit circle is 1, by definition!). cos(30) = (√3)/2 and sin(30) = 1/2 so this point can be represented by the complex value (√3)/2 + i(1/2).

Just work your way around the unit circle in a similar manner for the others.

.

Alan
Dec 18, 2014

#2**0 **

How do you go about finding all of these Alan

and are there no other numbers that would work?

Melody
Dec 18, 2014

#3**+10 **

Best Answer

They are the solutions to the 12th order polynomial x^{12} = 1

A 12th order polynomial has exactly 12 roots.

They can be pictured as equally spaced points on the unit circle in the complex plane.

Each root is separated from the next by an angle of 360/12 = 30 degrees.

Starting with the obvious root x = 1 + 0i on the x-axis, the next one (in a clockwise direction) has coordinates x = cos(30) and y = sin(30) (because the radius of the unit circle is 1, by definition!). cos(30) = (√3)/2 and sin(30) = 1/2 so this point can be represented by the complex value (√3)/2 + i(1/2).

Just work your way around the unit circle in a similar manner for the others.

.

Alan
Dec 18, 2014

#4**0 **

Thanks Alan :)

So if you are finding them all by hand it its going to be quite time consuming?

I knew there were supposed to be 12 answers but how can you prove that there aren't more?

Melody
Dec 19, 2014

#5**+5 **

Time consuming: Not really. Think about the symmetry. The off-axis ones are simply all the positive and negative combinations of 1/2 and √3/2. The on-axis ones are simply, 1, -1, i and -i.

As for proving there are exactly n roots to an n'th order polynomial, you'll have to go to Carl Friedrich Gauss (I think he was the first) for this!

.

Alan
Dec 19, 2014