Find all complex numbers satisfying z^6=1

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March 13th 2010, 01:35 PM
lucius

Find all complex numbers satisfying z^6=1

Hi guys.

"Find all complex numbers satisfying $z^6=1$ "

The hint is to factorise $z^6-1$ as a difference of squares.

$(z^3+1)(z^3-1)$
$z(z^2+\frac{1}{z}) \cdot z(z^2-\frac{1}{z})$

This seems incorrect.

The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.

I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the $cis\Theta$ where $\Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$ .

I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

I've asked several people about this problem, but they can't seem to distinguish copying from understanding.

I would love to really understand the solution to this problem.

Can you help?
March 13th 2010, 01:55 PM
e^(i*pi)

Quote:

Originally Posted by lucius

Hi guys.

"Find all complex numbers satisfying $z^6=1$ "

The hint is to factorise $z^6-1$ as a difference of squares.

$(z^3+1)(z^3-1)$
$z(z^2+\frac{1}{z}) \cdot z(z^2-\frac{1}{z})$

This seems incorrect.

The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.

I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the $cis\Theta$ where $\Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$ .

I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

I've asked several people about this problem, but they can't seem to distinguish copying from understanding.

I would love to really understand the solution to this problem.

Can you help?

I'd go with the sum/difference of two cubes afterwards.

$z^6-1=0$

$(z^3-1)(z^3+1)$

$z^3-1 = (z-1)(z^2+z+1)$

$z^3+1 = (z+1)(z^2-z+1)$

$z^6-1 = (z-1)(z+1)(z^2+z+1)(z^2-z+1)$

Use the quadratic formula to solve the last two terms which should give 6 solutions
March 13th 2010, 02:20 PM
Plato

Quote:

Originally Posted by lucius

"Find all complex numbers satisfying $z^6=1$ "
I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the $cis\Theta$ where $\Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$ .

I'm very interested in understanding the solution with De Moivre's theorem.

Define $\xi = e^{\frac{{i\pi }}{3}} = cis\left( {\frac{\pi }{3}} \right)$ .
Now the six roots are $\xi ^k ,\;k = 0,1,2, \cdots ,5$ .
March 13th 2010, 09:19 PM
kjf

math1051 assignment 1 @ UQ?
I'm stuck, too. I think i'm way out of my element here, but I just plugged in some values into de moivre's theorem.

I don't fully understand complex numbers, and I too am keen to know HOW to solve a question like this, rather than have someone say "here's the answer, off you go". Will be watching this thread.
March 13th 2010, 09:26 PM
Prove It

Isn't the answer provided by $e^{i\pi}$ clear enough?

It's simple application of factorisation rules DOTS and Sum/Difference of two cubes, and null factor law...

It's set equal to 0, so that means each factor could be 0. Solve for $z$ .
March 14th 2010, 08:41 PM
TheCoffeeMachine

Quote:

Originally Posted by lucius

I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

First, we express the RHS in the form | $z|(\cos\theta+i\sin\theta)$ , where $|z|$ is of course the modulus, and $\theta$ the argument. Since $|1| = 1$ , and $\arg{1} = 0$ , we have:

$1 = \cos(0)+i\sin(0) \Rightarrow z^6 = \cos(0)+i\sin(0) \Leftrightarrow z = \left\{cos(0)+i\sin(0)\right\}^{\frac{1}{6}}$

De Moivre's theorem states that: $\left\{\cos(\phi)+i\sin(\phi)\right\}^n = \cos{n}\phi+i\sin\phi$

Or more relevantly: $\left\{\cos(\phi)+i\sin(\phi)\right\}^{\frac{1}{n} }$ $= \cos\left[\dfrac{\phi+(n-1)2\pi}{n}\right]+i\sin\left[\dfrac{\phi+(n-1)2\pi}{n}\right]$

So $\left\{\cos(0)+i\sin(0)\right\}^{\frac{1}{n}}$ $= \cos\left[\dfrac{(n-1)2\pi}{6}\right]+i\sin\left[\dfrac{(n-1)2\pi}{6}\right]$

We just plug $n = 0, 1, 2,...5$ to get the six values and then convert them into the Cartesian form, i.e. $\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\p i}{3}\right) = \frac{1}{2}+\frac{\sqrt{3}}{2}i\$ .

Note that all the subsequent values after n = 5 are repetitions, i.e n = 0 has the same value as n = 6.
March 15th 2010, 10:42 PM
HakunaMatata

Quote:

Originally Posted by kjf

math1051 assignment 1 @ UQ?
I'm stuck, too. I think i'm way out of my element here, but I just plugged in some values into de moivre's theorem.

Haha yeah math1051. I did the same thing that was in the first resposnse, from e^(i*pi) - difference of 2 squares then differnces of two cubes. Its way easier that looking at D.M's theorem - way faster! and you still get the 6 solutions, so really, whats not to love? haha
March 15th 2010, 11:05 PM
Prove It

DeMoivre's Theorem isn't too bad as long as you remember that the roots are all evenly spaced around a circle.

So, in this case, remember that $z^6 = 1 = \cos{2\pi} + i\sin{2\pi}$ .

Therefore:

$z = \left[1\left(\cos{2\pi} + i\sin{2\pi}\right)\right]^{\frac{1}{6}}$

$= 1^{\frac{1}{6}}\left(\cos{\frac{2\pi}{6}} + i\sin{\frac{2\pi}{6}}\right)$

$= \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$ .

Since they are evenly spaced around the circle and there are 6 roots, that means they are all separated by an angle of $\frac{2\pi}{6} = \frac{\pi}{3}$ .

So the other roots are:

$\cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}}$

$\cos{\frac{3\pi}{3}} + i\sin{\frac{3\pi}{3}}$

$\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}$

$\cos{\frac{5\pi}{3}} + i\sin{\frac{4\pi}{3}}$

and

$\cos{\frac{6\pi}{3}} + i\sin{\frac{6\pi}{3}}$ .

If you convert each of these to cartesians, you should get the same answers as those found by factorising and using Null Factor Law.
March 16th 2010, 05:31 PM
kjf

I've now done it both ways (ie; diff of squars and DM theorum).
Both worked.

to anyone else doing the 1051 assignment, what in the hell is he asking for in question 10?

"Suppose y = cos(x), where 0<= x <= pi/2, obtain sin(x) as a function of y for the same interval.

Is he just asking, what sin function will give us the same "plot" as the cos function? something like 1-sin(x), i've done it, i just can't find it. But I don;t know if that's what we have to do?
March 16th 2010, 06:52 PM
Prove It

Quote:

Originally Posted by kjf

I've now done it both ways (ie; diff of squars and DM theorum).
Both worked.

to anyone else doing the 1051 assignment, what in the hell is he asking for in question 10?

"Suppose y = cos(x), where 0<= x <= pi/2, obtain sin(x) as a function of y for the same interval.

Is he just asking, what sin function will give us the same "plot" as the cos function? something like 1-sin(x), i've done it, i just can't find it. But I don;t know if that's what we have to do?

You should know that

$\sin^2{x} + \cos^2{x} = 1$

so

$\sin^2{x} = 1 - \cos^2{x}$

$\sin{x} = \pm \sqrt{1 - \cos^2{x}}$ .

Since you are only dealing with the first quadrant, you know that sine is positive, so

$\sin{x} = \sqrt{1 - \cos^2{x}}$

$= \sqrt{1 - y^2}$ .
March 17th 2010, 07:31 PM
kjf

Thankyou so much Prove It. I'm kind of a rope-learner, for things I have never fully encountered before I need to be shown how to do it. In fact, all I had to do was look at your post, remember than sin^2x + cos^2x = 1 and I had a go. compared my answer to yours and it came out the same!

Thanks again!