Solving Complex Roots for z^6 + z^3 + 1: Homework Help and Strategies

[ribbon]

Homework Statement

Find all complex solutions of z^6 + z^3 + 1

(z^3 + 1)/(z^3 - 1) = i

Homework Equations

The Attempt at a Solution

I am going crazy with trial and error with these, there must be some systematic method or tricks that I am oblivious of. For the second question I factored it to:

[(z+1)(z^2 - z + 1)]/[(z-1)(z^2 + z +1)] = i

obviously we have conditions z cannot equal 1. But also z cannot = -1/2 +- sqrt(3)i What happens here?

 

[micromass]

Substitute $u = z^3$ and solve for $u$.

 

[lurflurf]

You need to know about roots of unity.http://en.wikipedia.org/wiki/Root_of_unity
Do you know any trigonometry?
if so think about
cos(2p/3)+i*sin(2pi/3)

 

[ribbon]

Substitute $u = z^3$ and solve for $u$.

Wicked alright let u = z^3

We have u^2 + u + 1 = 0

then u = -1/2 +- [(√3)/2]i

Now given that we took u = z^3, can I just cube the results to get the answers for z??

 

[micromass]

Wicked alright let u = z^3

We have u^2 + u + 1 = 0

then u = -1/2 +- √3i

Now given that we took u = z^3, can I just cube the results to get the answers for z??

You know that

$$z^3 = -\frac{1}{2} \pm i \sqrt{3}$$

Then what is $z$?

 

[ribbon]

You know that

$$z^3 = -\frac{1}{2} \pm i \sqrt{3}$$

Then what is $z$?

OMG how silly of me yes you'd take the cube root of the positive and negative answers each so,

z = (-1 + i3^1/6)/(2^1/3)

z = (-1 - i3^1/6)/(2^1/3)

 

[ribbon]

However we would have 6 roots to this polynomial would we not, by the fundamental theorem of algebra?

 

[micromass]

I have no idea what you just did.

 

[ribbon]

I have no idea what you just did.

Oh no, do you get something different when you take the cube root?

 

[micromass]

Oh no, do you get something different when you take the cube root?

For any nonzero complex number, there are in fact three cube roots! Did you cover in class how to find these?

 

[ribbon]

For any nonzero complex number, there are in fact three cube roots! Did you cover in class how to find these?

No my book never mentioned anything about this, but it makes sense that we have a positive and a negative version of the root. That would give me four if I included the conjugates to the one I wrote, what would the 3rd set be??

 

[micromass]

Did you learn that you can write any complex number $z$ as

$$z = r(\cos(\theta) + i \sin(\theta))$$

Did you learn how to find $z^n$ using this representation?

 

[ribbon]

Did you learn that you can write any complex number $z$ as

$$z = r(\cos(\theta) + i \sin(\theta))$$

Did you learn how to find $z^n$ using this representation?

That kind of looks like the set up to DeMoivre theorem?

I know if it is r(cosθ + isinθ)^n = r^n(cos nθ + isin nθ)

But how does this help us?

 

[micromass]

Right, so we need to find $z = r(\cos(\theta) + i\sin(\theta))$ such that

$$r^3 (\cos(3\theta) + i\sin(3\theta) ) = \frac{1}{2} \pm i\sqrt{3}$$

So, try to write the RHS in the same $r(\cos(\theta) + i\sin(\theta))$ form, and see if you can find $r$ and $\theta$.

 

[ribbon]

Right, so we need to find $z = r(\cos(\theta) + i\sin(\theta))$ such that

$$r^3 (\cos(3\theta) + i\sin(3\theta) ) = \frac{1}{2} \pm i\sqrt{3}$$

So, try to write the RHS in the same $r(\cos(\theta) + i\sin(\theta))$ form, and see if you can find $r$ and $\theta$.

Ok we have cosine of an angle corresponding to 1/2 and the sin corresponding to (sqrt3)/2 so the
θ in question should be ∏/3 (60 degrees).

The r = (1/2)^2 + (sqrt3/4)^2
r = 1/4 + 3/4
r = 1

 

[micromass]

Ok we have cosine of an angle corresponding to 1/2 and the sin corresponding to (sqrt3)/2 so the
θ in question should be ∏/3 (60 degrees).

The r = (1/2)^2 + (sqrt3/4)^2
r = 1/4 + 3/4
r = 1

OK, so you need to find $r$ and $\theta$ such that

$$r^3(\cos(3\theta) + i \sin(3\theta)) = \cos(\pi/3) + i\sin(\pi/3)$$

This comes down to

$$r^3 = 1,~\cos(3\theta ) = \cos(\pi/3),~\sin(3\theta) = \sin(\pi/3)$$

Can you find $r$ and $\theta$ that satisfy this?

 

[ribbon]

OK, so you need to find $r$ and $\theta$ such that

$$r^3(\cos(3\theta) + i \sin(3\theta)) = \cos(\pi/3) + i\sin(\pi/3)$$

This comes down to

$$r^3 = 1,~\cos(3\theta ) = \cos(\pi/3),~\sin(3\theta) = \sin(\pi/3)$$

Can you find $r$ and $\theta$ that satisfy this?

Sure so r = 1 and θ=1pi/9

 

[micromass]

Sure so r = 1 and θ=1/9

Your $r$ is correct, your $\theta$ is not. There are multiple $\theta$ that solve this by the way.

 

[ribbon]

Your $r$ is correct, your $\theta$ is not. There are multiple $\theta$ that solve this by the way.

cos(pi/3) = cos(3θ)
cant I take cos inverse of both sides yielding

pi/3 = 3θ?

then θ=1/9 pi of course times some scalar k?

 

[micromass]

cos(pi/3) = cos(3θ)
cant I take cos inverse of both sides yielding

pi/3 = 3θ?

then θ=1/9 pi of course times some scalar k?

No, you can't just take the cos inverse of both sides. That way you'll lose solutions. However, $\pi/9$ is correct, but there are two other solutions.

When I ask you to solve $\cos(x) = \cos(y)$. How would you do it? Try to draw a graph if you don't see it.

 

[ribbon]

No, you can't just take the cos inverse of both sides. That way you'll lose solutions. However, $\pi/9$ is correct, but there are two other solutions.

When I ask you to solve $\cos(x) = \cos(y)$. How would you do it? Try to draw a graph if you don't see it.

Cosine is an even function so throw in -pi/9 as well.

As for the third how about 17∏/9?

 

[micromass]

Cosine is an even function so throw in -pi/9 as well.

As for the third how about 17∏/9?

Remember, it needs to satisfy $\sin(3\theta ) = \sin(\pi/3)$ as well.

 

[ribbon]

Remember, it needs to satisfy $\sin(3\theta ) = \sin(\pi/3)$ as well.

Ah yes so forget the negative ninth pi, was 17pi/9 correct? And 19pi/9?

 

[micromass]

Stop guessing. If you know $\cos(x) = \cos(y)$ and $\sin(x) = \sin(y)$, then what do you know about $x$ and $y$??

Well, certainly $x=y$ is a possibility. But so is $x=y + 2\pi$. Indeed, adding $2\pi$ doesn't change sine and cosine. Adding $4\pi$ doesn't change it either. Subtracting $2\pi$ doesn't change it either.

So we see that if $\cos(x) = \cos(y)$ and $\sin(x) = \sin(y)$, then $x = y + 2k\pi$ for some $k\in \mathbb{Z}$.

Now, apply this to your problem.