Solving Roots of Unity with De Moivre's Theorem

[h0llow]

Homework Statement

Use De Moivre's Theorem to solve for the roots of unity 1, ω, ω²
Hence show that the sum of these roots is zero

Homework Equations

r(cosθ + isinθ)
r(cos(θ + 2n∏)+isin(θ+isin∏))

The Attempt at a Solution

I know the first root,1, is 1(cos 0 + i sin 0)
but have no clue about the next 2, or of how i would prove they are equal to 0.

 

[Mark44]

Homework Statement

Use De Moivre's Theorem to solve for the roots of unity 1, ω, ω²
Hence show that the sum of these roots is zero

I don't see anywhere in your post that you are looking for the cube roots of 1.

Homework Equations

r(cosθ + isinθ)
r(cos(θ + 2n∏)+isin(θ+isin∏))

The Attempt at a Solution

I know the first root,1, is 1(cos 0 + i sin 0)
but have no clue about the next 2, or of how i would prove they are equal to 0.

The cube roots are not equal to 0; you are supposed to show (i.e., prove) that the sum of these three roots is zero.

If z represents one of these roots, what equation should you be trying to solve?

 

[h0llow]

r(cos(θ + 2n∏)+isin(θ+isin∏))?

 

[Mark44]

r(cos(θ + 2n∏)+isin(θ+isin∏))?

That's not an equation.

The idea that these are cube roots of unity is important.

 

[HallsofIvy]

Your problem said "use DeMoivre's theorem".

So, what is DeMoivre's theorem?

 

[h0llow]

(r(cosθ+isinθ))ⁿ

 

[Mark44]

(r(cosθ+isinθ))ⁿ

Equals what?

And in this problem, what's your best guess as to what n is?

 

[h0llow]

(r(cosθ+isinθ))ⁿ=rⁿe^inθ

hmm I am guessing for 1, n= 0
for ω, n=1
for ω², n = 2

 

[Mark44]

You're missing an important part. See http://en.wikipedia.org/wiki/De_Moivre's_formula.

When someone says "cube" what number comes to mind?

 

[h0llow]

well ^3, but I am not sure why it is significant.

 

[Mark44]

Because you're asked to find the three cube roots. That's why I have be emphasizing cube, since there are three dimensions in a cube.

 

[Mark44]

The (square, cube, fourth, fifth, etc.) roots of unity are spread out evenly around the unit circle. Does that suggest something about the angle between each pair of contiguous roots?

Aren't there any examples in your book that are similar to this problem? Hasn't some of this been covered in your class?

 

[h0llow]

how do you know they are cube roots?

Angle is the same for all values of n?(even integers)

EDIT:
no, this is the first time i have seen a question like this. The maths course in my country is currently changing(introduced in phases), so not everything is in books(99% is, but 1% like this aren't properly explained). And my finals start in 6 days =/.
I would normally ask my teacher, but he said he had no idea where to begin.

 

[Mark44]

how do you know they are cube roots?

Because they asked for three roots - 1, ω, and ω².

Use De Moivre's Theorem to solve for the roots of unity 1, ω, ω²

Angle is the same for all values of n?(even integers)

No. For different values of n, you get different angles. For a given n, what is the angle between consecutive roots?

Also, what do mean by "even integers"? The ones that are multiples of two? Or are you saying, "even for integers, the angle is the same"? Either way, I don't understand the question.

 

[h0llow]

okay, i understand the cube part.

Can you just tell me how you would go about solving it?

 

[Mark44]

Use De Moivre's Theorem to solve z³ = 1

Each solution is a cube root of unity (1).

 

[h0llow]

0_o

z³=1+0i = (1+0i)^1/3...r=1 θ=0

z1 = (1(cos0+isin0))^1/3
1(1+0)
=1

z2 = (1(cos2∏+isin2∏))^1/3
1(cos(2∏/3)+isin(2∏/3))
= -1/2 + √3/2

z3 = (1(cos(4∏)+isin(4∏))^1/3
=1(cos(4∏/3)+isin(4∏/3))
=-1/2 - √3/2



1+(-1/2+√3/2)+(-1/2 - √3/2) = 0!



Thank you so much!

Maths is amazing when u get it !


P.S. as you increase value of z (z..z1..z2) should θ increase by ∏ or 2∏?..here i used 2∏ but not sure.

 

[Mark44]

0_o

z³=1+0i = (1+0i)^1/3...r=1 θ=0

z1 = (1(cos0+isin0))^1/3
1(1+0)
=1

z2 = (1(cos2∏+isin2∏))^1/3
1(cos(2∏/3)+isin(2∏/3))
= -1/2 + √3/2

z3 = (1(cos(4∏)+isin(4∏))^1/3
=1(cos(4∏/3)+isin(4∏/3))
=-1/2 - √3/2



1+(-1/2+√3/2)+(-1/2 - √3/2) = 0!



Thank you so much!

Maths is amazing when u get it !

You're missing the imaginary parts on two of them.
z₂ = -1/2 + √3/2 * i
z₃ = -1/2 - √3/2 * i

P.S. as you increase value of z (z..z1..z2) should θ increase by ∏ or 2∏?..here i used 2∏ but not sure.

2$\pi$

 

[h0llow]

You're missing the imaginary parts on two of them.
z₂ = -1/2 + √3/2 * i
z₃ = -1/2 - √3/2 * i
2$\pi$

o you oops :P

thanks so much!... the word unity kind of confused me =/