Question

Can someone explain to me how I would find some of the roots? (problem posted below)

I know that I have to take x^12-1 and basically split it up. It then becomes (x^6-1) (x^6+1), and I could split the left one even further like this: (x^3-1) (x^3 +1). The roots I've collected so far are 1,-1,i,-i.

What I'm getting stuck on is finding the roots for (x^6 + 1). Can someone explain to me how I would go about finding them?

Answer 1

My professor said that I would need to use the unit circle to find them, but his explanation was rather vague.

Answer 2

I'm not sure how much you've learned about the unit circle when in the complex numbers, but there is an interesting and important relation between the polynomial equations \(x^n-1\) and the unit circle. Without getting too far into the details, basically every solution of \(x^n-1\) will be on the unit circle spaced evenly apart.

Answer 3

Yes, I did learn about that. It was described to me to visualize it as even slices out of the circle.

Answer 4

And, if I understand this correctly, since it is x^12, there should be 12 points evenly spaced around the unit circle.

Answer 5

That's exactly the case, and you shouldn't have too much trouble actually finding the solutions. The most difficult part will be writing them in the forms needed.

Answer 6

is \(z_{2\pi}\) a cyclic group of period 2pi ?

Answer 7

I think you'd use deMoirves?

Answer 8

Hmmm, ok . . . . like I said, I did find four of the solutions, it's just the rest that are giving me trouble.

Answer 9

@ikram002p, a cyclic group? I'm not sure I understand what you mean. @wio, I'm afraid I don't know what deMoirves is.

Answer 10

What forms of writing complex numbers have you learned about? In particular, what does \(Z_{2\pi}(t)\) mean?

Answer 11

Z[2π](t) means a complex number point (aka the terminal point) in terms of radians on the unit circle.

Answer 12

if you want to work it by factoring :-
x^6 + 1 = (x^2)^3 + 1^3 = (x^2+1)(x^4-x^2+1)

Answer 13

but clearly your prof wants you to use the unti circle/demoivre - NOT factoring !

Answer 14

Unfortunately, I believe that's what he wants, yes.

Answer 15

I'll admit, I'm unfamiliar with that particular notation. Could you give me an example of how it's used? How would you write \(i\) or \(i+1\) for example?

Answer 16

Umm, I would think you would know, at this stage in the game: \[
z^n = (re^{i\theta})^n=r^ne^{i\theta n}
\]

Answer 17

Is it that notation? I was afraid it might be different.

Answer 18

\[
=r^n[\cos(n\theta)+i\sin(n\theta)]
\]

Answer 19

Since we have:\[
1 = 1^6[\cos (6\theta) + i\sin(6\theta)]
\]We know the \(\cos\) term should go to \(1\) and the \(\sin\) term should go to \(0\). That is the only way to get a real number here.

Answer 20

Well, the example he gave the class was to find the roots of x^8 -1. First, we subtracted 1 and made it x^8=1.

We found the roots -1, 1, i, -i like I showed before. Then, he said to pick a random point on the unit circle and test it out. He picked (√2/2 + √2/2i). He then factored like this: √2/2(1 + i), and put in that value for x.

He said to start with squaring the value, so it became 2/4(2i). Which became 1/2(2i) which equals i. And that apparently equals 1 when you do the whole thing out.

Answer 21

@wio, I'm not familiar with what you just did with r and e . . .

Answer 22

So the method he gave you was guess and check?!

Answer 23

lol! :O

Answer 24

Hmm. Well, let's go back to dividing the unit circle up and worry about notation later. So you already know that you want to divide the unit circle up into 12 equal slices. The best way to do that (in my opinion) is to figure out what each angle will be. Since the entire unit circle is \(2\pi\) radians around, 1/12 of that will be \(2\pi/12=\pi/6\). So one of your solutions should be on the unit circle placed at \(\pi/6\) radians away from the x-axis.

To find the next solution, add \(\pi/6\) again to get \(2\pi/6=\pi/3\). One more time, and you get \(\pi/3+\pi/6=\pi/2\). This will at least get you the angle of all the solutions.

Answer 25

Pretty much, yea. I tried doing that myself, but nothing I seem to do with that works. And I'm sure he wants us to "guess" by picking a point that "looks" like it will work, but I don't how how I would be able to tell that.

Answer 26

if you assume they are evenly spaced for whatever bizarre reason.

Answer 27

We want twelfths, not sixths wio.

Have you ever seen/used the form \[r(\cos(\theta)+i\sin(\theta))\]for complex numbers?

Answer 28

I just sketched this out on some graph paper. Once I find the angle, perhaps I can use the terminal points of the angle to find another root?

Answer 29

have you learned

cos(x) + i sin(x) where x is an angle ?

your roots will start at x=0, and be equally spaced every 360/12 = 30º

Answer 30

\[
a = \cos(\pi/3)\\
bi = \sin(\pi/3)i
\]