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FP2 Complex Numbers Question

Hi, I've got a question which is the first part asks to show that
(z4eiθ)(z4eiθ)=z82z4cosθ+1(z^4-e^{i\theta})(z^4-e^{-i\theta})=z^8-2z^4cos\theta+1
Which I've managed to do, however, the second part asked to hence solve the equation
z8z41=0z^8-z^4-1=0
giving my answer in the form eiϕe^{i\phi} where π<ϕπ-\pi<\phi\leq\pi
Though I recognise the similarities, I'm really not sure how to go about doing this Would somebody be able to help me? Thank you!
(edited 8 years ago)
Reply 1
Original post by foorganders
Hi, I've got a question which is the first part asks to show that
(z4eiθ)(z4eiθ)=z82z4cosθ+1(z^4-e^{i\theta})(z^4-e^{-i\theta})=z^8-2z^4cos\theta+1
Which I've managed to do, however, the second part asked to hence solve the equation
z8z41=0z^8-z^4-1=0
giving my answer in the form eiϕe^{i\phi} where π<ϕπ-\pi<\phi\leq\pi
Though I recognise the similarities, I'm really not sure how to go about doing this Would somebody be able to help me? Thank you!


cosθ =1/2
θ /3

z4 = e^(iπ/4) and similarly the other


etc
First, what is your value of theta in the second polynomial (comparing it with the first)?

Seems TeeEm has already given you the answer, but as a second question, why pick θ=π/3 \theta = \pi/3 and not something like θ=7π3 \theta = \frac{7 \pi}{3} etc?

Original post by foorganders
Hi, I've got a question which is the first part asks to show that
(z4eiθ)(z4eiθ)=z82z4cosθ+1(z^4-e^{i\theta})(z^4-e^{-i\theta})=z^8-2z^4cos\theta+1
Which I've managed to do, however, the second part asked to hence solve the equation
z8z41=0z^8-z^4-1=0
giving my answer in the form eiϕe^{i\phi} where π<ϕπ-\pi<\phi\leq\pi
Though I recognise the similarities, I'm really not sure how to go about doing this Would somebody be able to help me? Thank you!
(edited 8 years ago)
Reply 3
Ah, thank you so much, I didn't realise it was as simple as that.

I imagine you would pick θ as π/3 and not 7π/3 or whatever because it must be between π and -π?
Precisely, the argument of a complex number is normally defined on the domain θ(π,π] \theta \in (-\pi,\pi]

Original post by foorganders
Ah, thank you so much, I didn't realise it was as simple as that.

I imagine you would pick θ as π/3 and not 7π/3 or whatever because it must be between π and -π?
Original post by WishingChaff
First, what is your value of theta in the second polynomial (comparing it with the first)?

Seems TeeEm has already given you the answer, but as a second question, why pick θ=π/3 \theta = \pi/3 and not something like θ=7π3 \theta = \frac{7 \pi}{3} etc?


Could you elaborate on why we can just freely choose θ\theta to make the RHS = The Octic ? I thought here theta is restricted to solutions of the Octic...?
Reply 6
Original post by SamKeene
Could you elaborate on why we can just freely choose θ\theta to make the RHS = The Octic ? I thought here theta is restricted to solutions of the Octic...?


When we choose θ=±π3\displaystyle \theta=\pm \frac{\pi}{3}

then this make RHS equal with LHS beacuse
Unparseable latex formula:

\dispplaystyle 2\cos \theta=1



so both side will be z8z4+1\displaystyle z^8-z^4+1

But in the second part of the question you have to solve the z8z41=0\displaystyle z^8-z^4-1=0 equation

for this substitute w=z4\displaystyle w=z^4 and solve the quadratic equation, then
find the 8 quartic roots and choose from them the solutions between -pi and pi
Here, theta is not related to the argument of the complex number z. The question labels the argument of the z variable using phi. Hence, theta is not restricted by the solutions to the polynomial.



Original post by SamKeene
Could you elaborate on why we can just freely choose θ\theta to make the RHS = The Octic ? I thought here theta is restricted to solutions of the Octic...?
Original post by WishingChaff
Here, theta is not related to the argument of the complex number z. The question labels the argument of the z variable using phi. Hence, theta is not restricted by the solutions to the polynomial.


I understand. Would it be correct to say that the solutions we arrive at require the necessary condition that cosθ=0.5cos{\theta}=0.5 to be met?

I.E the solution set is valid if and only if cosθ=0.5cos{\theta}=0.5.
They don't really require that cos(θ)=0.5 \cos(\theta) = 0.5 rather, you are able to compare the two forms of the polynomials and notice that they agree when that condition is satisfied. Hence, you are able to write the latter polynomial in a more useful form.

Original post by SamKeene
I understand. Would it be correct to say that the solutions we arrive at require the necessary condition that cosθ=0.5cos{\theta}=0.5 to be met?
Original post by WishingChaff
They don't really require that cos(θ)=0.5 \cos(\theta) = 0.5 rather, you are able to compare the two forms of the polynomials and notice that they agree when that condition is satisfied. Hence, you are able to write the latter polynomial in a more useful form.


I think whats throwing me of is TeeMe wrote:

cosθ =1/2
θ /3

z4 = e^(iπ/4) and similarly the other

Shouldn't the third line be:

z4 = e^(iπ/3) and similarly the other

I.e, the working might be:

(z4eiπ3)(z4eiπ3)=z8z4+1\displaystyle (z^4-e^{i\frac{\pi}{3}})(z^4-e^{-i\frac{\pi}{3}})=z^8-z^4+1

z4=eiπ3,z4=eiπ3\displaystyle z^4=e^{i\frac{\pi}{3}}, z^4=e^{-i\frac{\pi}{3}}

Edit but
z8z4+1z^8-z^4+1

Isn't the Octic we need to solve. So I am now confused again.
(edited 8 years ago)
Yeah, this looks correct.

Yes I noticed that the polynomial has a minus sign in. I believe this is a typo.

Original post by SamKeene
I think whats throwing me of is TeeMe wrote:

Shouldn't the third line be:

z4 = e^(iπ/3) and similarly the other

I.e, the working might be:

(z4eiπ3)(z4eiπ3)=z8z4+1\displaystyle (z^4-e^{i\frac{\pi}{3}})(z^4-e^{-i\frac{\pi}{3}})=z^8-z^4+1

z4=eiπ3,z4=eiπ3\displaystyle z^4=e^{i\frac{\pi}{3}}, z^4=e^{-i\frac{\pi}{3}}

Edit but
z8z4+1z^8-z^4+1

Isn't the Octic we need to solve. So I am now confused again.
(edited 8 years ago)
Original post by WishingChaff
Yeah, this looks correct.

Yes I noticed that the polynomial has a plus sign in. I believe this is a typo.


Thanks for the clarification, but check my edit. I'm confused again.

I can see how changing theta to get the LHS equal to:

z8z4+1z^8-z^4+1

Works, but we are asked to solved in the OP

z8z41z^8-z^4-1.

Is this an error or am I missing something?
Yeah, I think it is a typo.

Original post by SamKeene
Thanks for the clarification, but check my edit. I'm confused again.

I can see how changing theta to get the LHS equal to:

z8z4+1z^8-z^4+1

Works, but we are asked to solved in the OP

z8z41z^8-z^4-1.

Is this an error or am I missing something?
Original post by foorganders
...


Original post by WishingChaff
Yeah, I think it is a typo.


Ah good, thought I might be going crazy. I quoted the OP so he can clarify!
Good idea.

Original post by SamKeene
Ah good, thought I might be going crazy. I quoted the OP so he can clarify!
Hi, yeah I'm really sorry, that is a typo. The original question does indeed have +1 in the polynomial, not -1. Sorry for the confusion!
Original post by foorganders
Hi, yeah I'm really sorry, that is a typo. The original question does indeed have +1 in the polynomial, not -1. Sorry for the confusion!


Thank god, thought I was being real stupid earlier >.<

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