# Finding Complex Roots

1. Aug 5, 2013

### ribbon

1. The problem statement, all variables and given/known data
Find all complex solutions of z^6 + z^3 + 1

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

2. Relevant equations

3. 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?

2. Aug 5, 2013

### micromass

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

3. Aug 5, 2013

4. Aug 5, 2013

### ribbon

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??

5. Aug 5, 2013

### micromass

Staff Emeritus
You know that

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

Then what is $z$?

6. Aug 5, 2013

### ribbon

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)

Last edited: Aug 5, 2013
7. Aug 5, 2013

### ribbon

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

8. Aug 5, 2013

### micromass

Staff Emeritus
I have no idea what you just did.

9. Aug 5, 2013

### ribbon

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

10. Aug 5, 2013

### micromass

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

11. Aug 5, 2013

### ribbon

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??

12. Aug 5, 2013

### micromass

Staff Emeritus
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?

13. Aug 5, 2013

### ribbon

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?

14. Aug 5, 2013

### micromass

Staff Emeritus
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 $\theta$.

15. Aug 5, 2013

### ribbon

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

16. Aug 5, 2013

### micromass

Staff Emeritus
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?

17. Aug 5, 2013

### ribbon

Sure so r = 1 and θ=1pi/9

18. Aug 5, 2013

### micromass

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

19. Aug 5, 2013

### ribbon

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?

20. Aug 5, 2013

### micromass

Staff Emeritus
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.