The Student Room Group

Roots of unity

I'm very stuck on parts b and e.
Any help would be appreciated.
Question and what I've tried down below
Reply 1
16505698967843978442958228225389.jpg16505699429505520277217604090773.jpg16505699667638585314643102506377.jpg
Given that the answer manipulated into what I needed in c I then thought it would be a series sum but got nowhere with this as well

For part e I have no idea where to start
(edited 1 year ago)
Reply 2
In your answer to (b), I don't think you can assume that the z in this part is the same as the z in (a). In other words, I don't think you can replace the 1 by z5. But you have got to a point where you effectively have (something)5 = 1, so you could now use your answer to (a) to write (something) = (a list of different options that you've already worked out), and take it from there.
Reply 3
For (e), I think you need to look at the polynomial you vertified in (d) and apply your knowledge of what it means for it to have a z2 coeffiecient of 10, in terms of the roots of the polynomial.
Reply 4
Original post by Pangol
In your answer to (b), I don't think you can assume that the z in this part is the same as the z in (a). In other words, I don't think you can replace the 1 by z5. But you have got to a point where you effectively have (something)5 = 1, so you could now use your answer to (a) to write (something) = (a list of different options that you've already worked out), and take it from there.

Ok, i've got to the required solution by saying that (z-1)^5/(z+1)^5=1
so (z-1)/(z+1)=w^k
z=(w^k+1)/(1-w^k) which is correct
however what i don't understand is you don't reach the same conclusion if you start with (z+1)^5/(z-1)^5=1 and im not sure why?
Edit: just looked at it a bit more and is it because having (z-1) as the denominator could run into problems due to one of the roots being 1, so dividing by 0?
(edited 1 year ago)
Reply 5
Original post by Skiwi
Ok, i've got to the required solution by saying that (z-1)^5/(z+1)^5=1
so (z-1)/(z+1)=w^k
z=(w^k+1)/(1-w^k) which is correct
however what i don't understand is you don't reach the same conclusion if you start with (z+1)^5/(z-1)^5=1 and im not sure why?
Edit: just looked at it a bit more and is it because having (z-1) as the denominator could run into problems due to one of the roots being 1, so dividing by 0?

That seems likely to be it. I hadn't actually done it myself, but if it has got you the answers you were expecting, it's got to be the way to go.
Reply 6
Original post by Pangol
For (e), I think you need to look at the polynomial you vertified in (d) and apply your knowledge of what it means for it to have a z2 coeffiecient of 10, in terms of the roots of the polynomial.

Do you mind expanding on this slightly, im not too sure what to do?
Original post by Skiwi
Ok, i've got to the required solution by saying that (z-1)^5/(z+1)^5=1
so (z-1)/(z+1)=w^k
z=(w^k+1)/(1-w^k) which is correct
however what i don't understand is you don't reach the same conclusion if you start with (z+1)^5/(z-1)^5=1 and im not sure why?
Edit: just looked at it a bit more and is it because having (z-1) as the denominator could run into problems due to one of the roots being 1, so dividing by 0?

Agree about the approach, but note you'll only have 4 solutions (on the imaginary axis) and the divide by 0 is still a problem when you have
z=(w^k+1)/(1-w^k)
with k=0
(edited 1 year ago)
Reply 8
Original post by mqb2766
Agree about the approach, but note you'll only have 4 solutions (on the imaginary axis) and the divide by 0 is still a problem when you have
z=(w^k+1)/(1-w^k)

The next part of the question introduces a restriction of k having to be 1,2,3,4 which i think removes the problems
Original post by Skiwi
The next part of the question introduces a restriction of k having to be 1,2,3,4 which i think removes the problems

Agree, the problem is w=1, so k=0, but the answer to b) is 4 values. The other representation should work, why do you think it doesnt?
(edited 1 year ago)
Reply 10
Original post by mqb2766
Agree, the problem is w=1, so k=0, but the answer to b) is 4 values.

So i list the expression of z 4 times replacing k with 1,2,3,4 each time?
Original post by Skiwi
Do you mind expanding on this slightly, im not too sure what to do?

It's an idea from the roots of polynomials topic. You should know things about certain combinations of the roots of a polynomial and the coefficients of that polynomial. (Sorry this is vague, don't want to spell it out too much!)
Reply 12
Original post by Pangol
It's an idea from the roots of polynomials topic. You should know things about certain combinations of the roots of a polynomial and the coefficients of that polynomial. (Sorry this is vague, don't want to spell it out too much!)

I think i'm somewhat on the right lines,
what i've done is:
The equation has 4 roots w,w^2,w^3,w^4
we know from roots of polynomials that ab+ay+ad+by+bd+yd=2
so w^3+w^4+2w^5+w^6+w^7=2
w+w^2+w^3+w^4=0
a bit stuck on how to progress as this is clearly wrong as they should add to -1
Reply 13
@Pangol @mqb2766 thanks for the help, i finally managed to show that is was equal to 2.

5z^4+10z^2+1=0 has 4 roots, the ones found in part c.
if we let y=z^2 then we get 5y^2+10y+1=0
this quadratic has 2 roots, (icotpi/5)^2 and (icot2pi/5)^2. ( i picked these because it's what the question required but you could use the 3pi and 4pi terms as well due to the fact that the pi/5 and 4pi/5 square to the same value, ditto for 2pi/5 and 3pi/5)
from roots of polynomials alpha+beta=-b/a
(icotpi/5)^2+(icot2pi/5)^2=-2
cot^2(pi/5)+cot^2(2pi/5)=2 as needed.

If i've missed anything/made an assumption or i've done it in a slightly harder way please tell me.
Sounds like you're there / nearly there. Did you get it sorted last night why
(z-1)/(z+1)=w^k => z = (1+w^k)/(1-w^k)
(z+1)/(z-1)=w^k => z = -(1+w^k)/(1-w^k)
give the same result (k=1..4) for b)? k=0 is not a solution in both cases.

For e) going down your original approach, though its z, not w:
- cot(pi/5)cot(2pi/5) - cot(pi/5)cot(3pi/5) - cot(pi/5)cot(4pi/5) - cot(2pi/5)cot(4pi/5) - cot(2pi/5)cot(3pi/5) - cot(3pi/5)cot(4pi/5) = 2
Then just use cot(x) = -cot(pi-x) supplementary identity and stuff cancels. It follows directly on from c) and d).

Note here that the transformation (z+1)/(z-1) maps a circle (root of unity) to a line (imaginary axis). Its a simple Mobius transformation.
(edited 1 year ago)
Reply 15
Original post by mqb2766
Sounds like you're there / nearly there. Did you get it sorted last night why
(z-1)/(z+1)=w^k => z = (1+w^k)/(1-w^k)
(z+1)/(z-1)=w^k => z = -(1+w^k)/(1-w^k)
give the same result (k=1..4) for b)? k=0 is not a solution in both cases.

For e) going down your original approach, though its z, not w:
- cot(pi/5)cot(2pi/5) - cot(pi/5)cot(3pi/5) - cot(pi/5)cot(4pi/5) - cot(2pi/5)cot(4pi/5) - cot(2pi/5)cot(3pi/5) - cot(3pi/5)cot(4pi/5) = 2
Then just use cot(x) = -cot(pi-x) supplementary identity and stuff cancels. It follows directly on from c) and d).

Note here that the transformation (z+1)/(z-1) maps a circle (root of unity) to a line (imaginary axis). Its a simple Mobius transformation.

I get the bit for part b now. I like your method for part e a lot more than how I approached it, it's solely algebraic and doesn't rely on the calculator to show that the squared roots give the 4 original roots.
Although I have no idea what a mobius transformation is.
Original post by Skiwi
I get the bit for part b now. I like your method for part e a lot more than how I approached it, it's solely algebraic and doesn't rely on the calculator to show that the squared roots give the 4 original roots.
Although I have no idea what a mobius transformation is.

The part e) pretty much followed on from the previous parts and you had the right idea, but seemed to use the roots for w not z. The question was leading you through the map from (points on) the unit circle (w space) to (points on) the imaginary axis (z space) using the w = (z+1)/(z-1) map.

The mobius stuff was for you to google if youre interested there are a few videos etc on it. Its beyond a level, but its a fairly famous complex map and some of the visualizations are intereting. For this mapping
(z+1)/(z-1) = 1 + 2/(z-1) = w
If you think about the complex planes for w and z. The origin in w (0,0) is mapped to z (-1,0). The unit circle in w is mapped to the imaginary axis in z (this question). The positive real axis in z is mapped to the real axis segment (-1,1) in w etc. But w = (1,0) (again this question) has no solution (mapped to infinity in z space).

Its obviously beyond whats necessary for the question though.
(edited 1 year ago)
Reply 17
Original post by mqb2766
The part e) pretty much followed on from the previous parts and you had the right idea, but seemed to use the roots for w not z. The question was leading you through the map from (points on) the unit circle (w space) to (points on) the imaginary axis (z space) using the w = (z+1)/(z-1) map.

The mobius stuff was for you to google if youre interested there are a few videos etc on it. Its beyond a level, but its a fairly famous complex map and some of the visualizations are intereting. For this mapping
(z+1)/(z-1) = 1 + 2/(z-1) = w
If you think about the complex planes for w and z. The origin in w (0,0) is mapped to z (-1,0). The unit circle in w is mapped to the imaginary axis in z (this question). The positive real axis in z is mapped to the real axis segment (-1,1) in w etc. But w = (1,0) (again this question) has no solution (mapped to infinity in z space).

Its obviously beyond whats necessary for the question though.

Ok thanks a lot.

I'll pretend i understood half of what you said beyond the first paragraph. I'll try to find some videos and come back later to see if it makes sense then.

Quick Reply