psycholeo
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Let p = w + w^4 and q = w^2 + w^3, where w = e^2*pi*i/5. Show that p + q = -1 and and pq = -1.

Okay, so my workings out are as follows:

p + q = (w + w^4) + w^2 + w^3
= w + w^4 + w(w + w^2)
= w + w^4 - w
= w^4

w^4 = e^8*pi*i/5

plug x = 8pi/5 into (cos x + i sin x), expecting to get an answer of -1

nope, 0.30902 - 0.95106i.

Where have I gone wrong?
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RDKGames
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(Original post by psycholeo)
Let p = w + w^4 and q = w^2 + w^3, where w = e^2*pi*i/5. Show that p + q = -1 and and pq = -1.

Okay, so my workings out are as follows:

p + q = (w + w^4) + w^2 + w^3
= w + w^4 + w(w + w^2)
= w + w^4 - w
= w^4

w^4 = e^8*pi*i/5

plug x = 8pi/5 into (cos x + i sin x), expecting to get an answer of -1

nope, 0.30902 - 0.95106i.

Where have I gone wrong?
How do you know that w^2 + w = -1 ?? (Going from second line of working to the third)
This is unjustified.

Instead, use the fact that w^4 + w^3 + w^2 + w + 1 = \dfrac{w^5 -1}{w-1}
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psycholeo
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(Original post by RDKGames)
How do you know that w^2 + w = -1 ?? (Going from second line of working to the third)
This is unjustified.

Instead, use the fact that w^4 + w^3 + w^2 + w + 1 = \dfrac{w^5 -1}{w-1}
I thought that what true? Or is that just for cube roots of unity?
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RDKGames
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(Original post by psycholeo)
I thought that what true? Or is that just for cube roots of unity?
Yes only for cube roots.

If w^3 = 1 then w^3 -1 =0 hence \dfrac{w^3-1}{w-1} = 0 hence w^2 + w + 1 =0 hence w^2 + w = -1.
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psycholeo
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Sorry, how do you know that [img=273x45]https://www.thestudentroom.co.uk/latexrender/pictures/6a/6aa76ae7212129196979be5d4bcc8bdd.png[/img]

And where does the 1 come from?
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RDKGames
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(Original post by psycholeo)
Sorry, how do you know that Image

And where does the 1 come from?
Well to me at A-Level it was given and proven that \dfrac{w^{n+1} -1 }{w-1} = w^{n} + w^{n-1} + \ldots + w + 1 so I would expect you to have covered it??

Anyhow, if you're not comfortable with this result, I suggest you learn it and its proof.

https://math.stackexchange.com/quest...x-1xn-1xn-2-x1
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psycholeo
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(Original post by RDKGames)
Well to me at A-Level it was given and proven that \dfrac{w^{n+1} -1 }{w-1} = w^{n} + w^{n-1} + \ldots + w + 1 so I would expect you to have covered it??

Anyhow, if you're not comfortable with this result, I suggest you learn it and its proof.

https://math.stackexchange.com/quest...x-1xn-1xn-2-x1
Are there other ways to show it without using that? That's not taught to us and where did you get the +1 from when p + q = w + w^2 + w^3 + w^4
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RDKGames
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(Original post by psycholeo)
Are there other ways to show it without using that? That's not taught to us and where did you get the +1 from when p + q = w + w^2 + w^3 + w^4
Just rewrite it as p+q = (w^4+w^3+w^2+w+1) -1 then.

I'm fairly sure this is the way they'd want you to do it. What is your exam board? I can't see a different nice way about it.
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psycholeo
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(Original post by RDKGames)
Just rewrite it as p+q = (w^4+w^3+w^2+w+1) -1 then.

I'm fairly sure this is the way they'd want you to do it. What is your exam board? I can't see a different nice way about it.
But surely it's show so you can't use p + q = -1 to show p + q = -1. And it's from NEAB June 1998
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RDKGames
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(Original post by psycholeo)
But surely it's show so you can't use p + q = -1 to show p + q = -1. And it's from NEAB June 1998
We are not using the fact to show itself is true... that makes no logical sense. Where are you getting that impression??

We begin with p+q in terms of w and proeed with logical steps to get down to -1.

If you do not want to use my approach, then you can write out:

$\begin{align*} p+q & = w^4 + w^3 + w^2 + w \\ & = [\cos(\frac{8\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{2\pi}{5})] + i[\sin(\frac{8\pi}{5}) +  \sin(\frac{6\pi i}{5}) + \sin(\frac{4\pi}{5}) + \sin(\frac{2\pi}{5})] \\ & = [\cos(-\frac{2\pi}{5}) + \cos(-\frac{4\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{2\pi}{5})] + i \underbrace{[\sin(-\frac{2\pi}{5}) +  \sin(-\frac{4\pi}{5}) + \sin(\frac{4\pi}{5}) + \sin(\frac{2\pi}{5})]}_{=0 \ \text{since} \ \sin(-x) \equiv -\sin x} \\ & = 2\cos(\frac{2\pi}{5}) + 2\cos(\frac{4\pi}{5}) \end{align*}$

and simplify from there.
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psycholeo
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Are you sure there's no other way without using the formula you have given?
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RDKGames
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(Original post by psycholeo)
Are you sure there's no other way without using the formula you have given?
Look at my edit.
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psycholeo
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(Original post by RDKGames)
Look at my edit.
Thank you for the help (even though you help with all of my student room threads haha). Out of interest once you've put the ws into cos and i sin form, as shown in the first line of your workings. Couldn't you just put it in a calculator and show that equals -1
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RDKGames
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(Original post by psycholeo)
Thank you for the help (even though you help with all of my student room threads haha). Out of interest once you've put the ws into cos and i sin form, as shown in the first line of your workings. Couldn't you just put it in a calculator and show that equals -1
Dunno, check the mark scheme. If they're posing a question like this when I'd imagine they want you to do a bit more than to just put something into a calculator.
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psycholeo
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(Original post by RDKGames)
Dunno, check the mark scheme. If they're posing a question like this when I'd imagine they want you to do a bit more than to just put something into a calculator.
Test isn't readily available as it's so far back and the textbook just has the answers. I'll just go with it.
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username3555092
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Knowledge of geometric series is unnecessary.

Note that 1=w^5.

So

\displaystyle (1)+w+w^2+w^3+w^4 &= (w^5)+w+w^2+w^3+w^4

You can use this to deduce something about 1+w+w^2+w^3+w^4.

For the second bit about pq, the expansion is quick and easy and you'll get four terms. But the line of thinking that might be helpful from there is something like:

w^6=w^5 \times w = w.
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