Is DeMoivres Theorem the way to go?

  • Thread starter thomas49th
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Homework Statement


Solve

[tex]z^{6} + z^{3} + 1 =0[/tex]


Homework Equations


DeMoivres


The Attempt at a Solution


Well I could apply DeMoivres right away, or would it be better to do this:

z³(z³+1) = -1

Does that mean I have z³ = -1 and z³ = -2. Am I allowed to say that? I don't feel to happy about the second term?

Thanks
Thomas
 

Answers and Replies

  • #2
fzero
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I would make a substitution that leads to an easier equation and then use de Moivre.
 
  • #3
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I would make a substitution that leads to an easier equation and then use de Moivre.

A-ha (good band)! u = z³.

But is there really a need to use De Moivre now? I just used the quadratice formula, turned it into polar form and got

1.366, -0.366 (are they right)?

Thanks
Thomas
 
  • #4
fzero
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A-ha (good band)! u = z³.

But is there really a need to use De Moivre now? I just used the quadratice formula, turned it into polar form and got

1.366, -0.366 (are they right)?

Thanks
Thomas

I wouldn't convert those to decimals, you can find an exact answer by leaving the radicals in there. The solutions of the original equation are the cube roots of those values, so you probably want to use de Moivre to find those.
 
  • #5
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I wouldn't convert those to decimals, you can find an exact answer by leaving the radicals in there. The solutions of the original equation are the cube roots of those values, so you probably want to use de Moivre to find those.

What do you mean by radicals?

Ahh I forgot to un-substitute:

u = z³

[tex]z = u ^{\frac{1}{3}}[/tex]. Now we can use De Moivre. Wheyyyy

[tex]z = u ^{\frac{1}{3}} = cos\frac{\pi}{9} \pm \sin \frac{\pi}{9} [/tex]

Is that my final answer?
Thanks
Thomas
 
  • #6
fzero
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I'm getting slightly different numbers, you might want to recheck your algebra or at least post more of your work. You seem to be ignoring factors of i in many places.

The quadratic solutions are

[tex]z^3 = \frac{-1\pm i\sqrt{3}}{2} = e^{\pm 2\pi i/3}.[/tex]

Next you'll have to apply de Moivre's theorem correctly to this. Remember that you should find 6 distinct roots in all.
 
  • #7
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I'm getting slightly different numbers, you might want to recheck your algebra or at least post more of your work. You seem to be ignoring factors of i in many places.

The quadratic solutions are

[tex]z^3 = \frac{-1\pm i\sqrt{3}}{2} = e^{\pm 2\pi i/3}.[/tex]

Next you'll have to apply de Moivre's theorem correctly to this. Remember that you should find 6 distinct roots in all.

where did your -ve sign for the 1 come from? I'm also getting
[tex]e^{\frac{\pi}{3}i}[/tex]. Theta is equal to arctan(sqrt(3)) = pi/3. Where is your 2 coming from?

6 roots because the original has a degree of 6 and that pi/3 goes into 2pi 6 times? We not concerned with negative pi? [tex]e^{\frac{k\pi}{3}i}[/tex] k =1,2,3,4,5,6

2.85,8.12,23.14,65.94,187.91,535.49

probably wrong...
 
  • #8
fzero
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where did your -ve sign for the 1 come from?

From the quadratic formula.

I'm also getting
[tex]e^{\frac{\pi}{3}i}[/tex]. Theta is equal to arctan(sqrt(3)) = pi/3. Where is your 2 coming from?

We want

[tex] \cos\theta = - \frac{1}{2},~\sin\theta = \pm \frac{\sqrt{3}}{2}[/tex]

so [tex]\theta = \pi/6 + \pi/2 = 2\pi/3[/tex].

We can rewrite [tex]-2\pi/3[/tex] as

[tex]2\pi - 2\pi/3 = 4\pi/3.[/tex]

6 roots because the original has a degree of 6 and that pi/3 goes into 2pi 6 times? We not concerned with negative pi? [tex]e^{\frac{k\pi}{3}i}[/tex] k =1,2,3,4,5,6

2.85,8.12,23.14,65.94,187.91,535.49

probably wrong...

Almost. Recheck your calculations for the roots of the quadratic.
 
  • #9
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Ahh the negative sign. Ofcourse.

[tex]
\cos\theta = - \frac{1}{2},~\sin\theta = \pm \frac{\sqrt{3}}{2}
[/tex]

How can you equate the real part of cos(x) and the imaginary to sin(x) like that? The theta should be equal for both parts shouldn't it? When you have r(cos(x) + isin(x)) you don't have 2 different x's.

[tex]
\theta = \pi/6 + \pi/2 = 2\pi/3
[/tex]

Where did these angles come from. Why are you adding them.

Maybe I'm too tired. Just not getting this.

In an earlier post you mentioned radicals. What do you mean?
 
  • #10
fzero
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Ahh the negative sign. Ofcourse.

[tex]
\cos\theta = - \frac{1}{2},~\sin\theta = \pm \frac{\sqrt{3}}{2}
[/tex]

How can you equate the real part of cos(x) and the imaginary to sin(x) like that? The theta should be equal for both parts shouldn't it? When you have r(cos(x) + isin(x)) you don't have 2 different x's.

I've already used that the modulus of the roots is 1, so I'm identifying [tex]\theta[/tex] by looking at the real and imaginary parts. If the roots are

[tex] x \pm i y = \cos \theta \pm i \sin\theta .[/tex]

I could have equally as well used the tangent, but this made it easier to see what quadrant we were in. There's only one [tex]\theta[/tex].

[tex]
\theta = \pi/6 + \pi/2 = 2\pi/3
[/tex]

Where did these angles come from. Why are you adding them.

From the signs above we see that [tex]\theta[/tex] is a multiple of [tex]\pi/6[/tex]. [tex]\pi/2[/tex] is the rotation into the 2nd quadrant.

Maybe I'm too tired. Just not getting this.

In an earlier post you mentioned radicals. What do you mean?

I meant the [tex]\sqrt{3}[/tex] in the quadratic roots. I didn't realize at the time that you were giving the arguments. The point still remains, if you have an exact expression in terms of [tex]\pi[/tex] (or [tex]\sqrt{3}[/tex]) there's no reason to convert to decimal expressions unless you have some other reason for it, like a physical computation.
 

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