Complex Root of Unity questions!! (1 Viewer)

[ledamn]

Anyone would like to try to work out and explain these root of unity questions?
1. Let w, a complex number, be the cube root of unity.
(a) find possible values of w.
(b) Show w^2 + w + 1 = 0
(c) Hence simplify, (1+w)^6

2. Given w, a complex number, is a root of equation z^3-1=0, find:
(a) a+bw+cw^2 / c + aw + bw^2

(b) (1-w+w^2)(1+w-w^2)

3. A polynomial R(z) is given by R(z) = z^6 -1. Let (alpha) not = to 1, be that complex root of R(z) =0 which has the smallest positive argument. Show that 1+(alpha)+(alpha)^2 + (alpha)^3....(alpha)^5 = 0.

 

[Axio]

Anyone would like to try to work out and explain these root of unity questions?
1. Let w, a complex number, be the cube root of unity.
(a) find possible values of w.
(b) Show w^2 + w + 1 = 0
(c) Hence simplify, (1+w)^6

2. Given w, a complex number, is a root of equation z^3-1=0, find:
(a) a+bw+cw^2 / c + aw + bw^2

(b) (1-w+w^2)(1+w-w^2)

3. A polynomial R(z) is given by R(z) = z^6 -1. Let (alpha) not = to 1, be that complex root of R(z) =0 which has the smallest positive argument. Show that 1+(alpha)+(alpha)^2 + (alpha)^3....(alpha)^5 = 0.

I've done a few. Some may be wrong.

1a. w is a cube root of unity, making w^3 =1. Clearly w=1 is a solution. The roots are equally spaced around the unit circle. Making w=1, cis2pi/3, cis-2pi/3= cos2pi/3 - sin2pi/3.

1b. w^2, w and 1 are cube roots of unity. Therefore w^2 + w + 1 = 2cos2pi/3 + sin2pi/3 - sin2pi/3 +1 = -1 +1 = 0

1c. w+1=-w^2. Therefore (w+1)^6 = (-w^2)^6=w^12 =cis8pi=1

2b. (1-w+w^2)(1+w-w^2)=(-2w)(-2w^2)=-2(w^3)=-2cis2pi=-2

 

[wiliaamnguuyen]

Hey Axio, could you explain how in 1c. How, w^12 = cis8pi = 1?

 

[Axio]

Hey Axio, could you explain how in 1c. How, w^12 = cis8pi = 1?

I went w^12=cis8pi (de Moivre). Then cis8pi= cis1440=cis360=1+0i=1

 

[wiliaamnguuyen]

Oh did you sub in roots for w? W has 3 roots, which one did you use exactly?

 

[wiliaamnguuyen]

And for 1b.
Can't you just show w^2 + w + 1=0
Cant you just factorise w^3-1=0?
(w-1)(w^2+w+1)=0
therefore, w^2 +w+1=0

 

[Axio]

Oh did you sub in roots for w? W has 3 roots, which one did you use exactly?

I used cis2pi/3. But if you think about it... w^12 =(w^3)^4=1^4=1.

 

[integral95]

For 2a

 

[integral95]

for 3

I'd use sum of the roots of the polynomial

 

[wiliaamnguuyen]

Nice nice bro!

 

[wiliaamnguuyen]

Hey @integral95, how would you use S.O.R in question 3?

 

[integral95]

Hey @integral95, how would you use S.O.R in question 3?

Yeah when you work out the roots of z^6-1 = 0 you pretty much get

1, (alpha), (alpha)^2 ,(alpha)^3....(alpha)^5 as the roots where alpha is a complex root with the smallest positive argument

use S.O.R of the polynomial which is 1+(alpha)+(alpha)^2 + (alpha)^3....(alpha)^5 =-0/1 = 0

 

[wiliaamnguuyen]

Oh I understand. Can you solve this hard one?

4. w is a complex cube root of 1. Show w^2 is also a root. Find the value of 1+w+w^2. Hence find the value of (1+2w+3w^3)(1+2w^2+3w).
5. Show solutions of equation (z+1)^4=4(z-1)^4 are 1+-2i and 1/5(1+-2i)
6. Find the quadratic equation, with integer coefficient that has roots 4+w, 4+w^2
7. Given (2+3i)^3=-46+9i, find another value of z such that z^3 = -46 + 9i

Gl!

 

[wiliaamnguuyen]

Nah for question 7, I kinda think I should change the question to make it seem more a root of unity question.

They give you (2+3i)^3 = -46 +9i & z^3 = -46 +9i ------> which both looks the same (-46+9i. Therefore

z^3= -46 +9i (sub in (2+3i)^3 = -46+9i)
z^3 = (2+3i)^3 (Both have the power same power. Now divide?)
(z/ 2+3i ) ^3 = 1

Sub w = (z/ 2+3i)^3
therefore,
w^3 = 1 (Root of unity question now?
Sub in w= rcis(theta)

r^3cis(3-theta) = cis(0) (Sub w= rcis theta, then use demoivres. Change the right hand side to mod/arg form)

r^3 = 1 3(theta) = 0 + k360
r=1 theta = k360/3
Where k = 0, +-1, +-2 ... etc.
k=0, theta= 0 ---> w = cis0 = 1
k=1, theta= 120 ----> w = cis3pi/2
k=2, theta = 240 -----> w= cis 3pi/4

Now I think sub w = (z / 2 + 3i)^3 because in question 7 asks, "to find other values of z" therefore you gotta sub the 3 'w' values you had just worked out above?
But, once you sub in w, to find z. It's too messy, so before subbing any 'w' values, I want to manipulate the given equation " w= (z / 2+3i)^3 " to make z the subject.

After making z the subject, then just sub in w to find z. Just algebra afterwards

 

[integral95]

Oh I understand. Can you solve this hard one?

4. w is a complex cube root of 1. Show w^2 is also a root. Find the value of 1+w+w^2. Hence find the value of (1+2w+3w^3)(1+2w^2+3w).
5. Show solutions of equation (z+1)^4=4(z-1)^4 are 1+-2i and 1/5(1+-2i)
6. Find the quadratic equation, with integer coefficient that has roots 4+w, 4+w^2
7. Given (2+3i)^3=-46+9i, find another value of z such that z^3 = -46 + 9i

Gl!

4. expand the whole thing and simply with w^3 = 1 and 1+w+w^2 = 0

5.just sub them in

6.x^2-(4+w+4+w^2)x+(4+w)(4+w^2)

 

[HeroicPandas]

4. expand the whole thing and simply with w^3 = 1 and 1+w+w^2 = 0

expanding seems tedious.. how about using the two things you gave us (w^3 = 1 and 1+w+w^2 = 0) straight away to simplify that mess so that expanding is easier?

 

[wiliaamnguuyen]

How would you simplify straight away using (w^3 = 1 and 1+w+w^2=0) to make expanding easier?

 

[integral95]

How would you simplify straight away using (w^3 = 1 and 1+w+w^2=0) to make expanding easier?

(1+2w+3w^3)(1+2w^2+3w) = (1+2w+3)(1+w+w^2+2w+w^2) = (2)(w+2)(2w+w^2) = (2w)(w+2)^2

 

[HeroicPandas]

w^3 = 1 .... (1)

1+w+w^2 = 0 ....... (2)

Simplify (showing important steps)
= (1+2w+3w^3)(1+2w^2+3w)

= (1 +2w + 3)(1+ 3w + 2w^2) (using ..(1) and slightly re-arranging second product)

= 2(2 + w)( 2 + 2w + 2w^(2) - 1 + w ) (using ...2 x (2))

= 2(2 + w)( 2(0) + w - 1 )

= 2(w + 2)(w - 1)

= 2(w^2 + w - 2)

= 2( -1 - 2 ) (using ...(2) and re-arranging, i.e. 1+w+w^2 = 0 -----> w+w^2 = -1)

= -6