# Thread: Find all complex numbers satisfying z^6=1

1. ## Find all complex numbers satisfying z^6=1

Hi guys.

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

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

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

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 $\displaystyle cis\Theta$ where $\displaystyle \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$.

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 would love to really understand the solution to this problem.

Can you help?

2. Originally Posted by lucius
Hi guys.

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

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

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

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 $\displaystyle cis\Theta$ where $\displaystyle \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$.

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 would love to really understand the solution to this problem.

Can you help?
I'd go with the sum/difference of two cubes afterwards.

$\displaystyle z^6-1=0$

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

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

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

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

Use the quadratic formula to solve the last two terms which should give 6 solutions

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

I'm very interested in understanding the solution with De Moivre's theorem.
Define $\displaystyle \xi = e^{\frac{{i\pi }}{3}} = cis\left( {\frac{\pi }{3}} \right)$.
Now the six roots are $\displaystyle \xi ^k ,\;k = 0,1,2, \cdots ,5$.

4. 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.

5. Isn't the answer provided by $\displaystyle e^{i\pi}$ 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 $\displaystyle z$.

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

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

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

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

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

We just plug $\displaystyle n = 0, 1, 2,...5$ to get the six values and then convert them into the Cartesian form, i.e. $\displaystyle \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.

7. 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

8. 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 $\displaystyle z^6 = 1 = \cos{2\pi} + i\sin{2\pi}$.

Therefore:

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

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

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

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

So the other roots are:

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

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

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

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

and

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

If you convert each of these to cartesians, you should get the same answers as those found by factorising and using Null Factor Law.

9. 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?

10. 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

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

so

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

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

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

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

$\displaystyle = \sqrt{1 - y^2}$.

11. 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!

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