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[M2] Please help complex number question

Please help complex number question

I attached my solutions for (i), (ii) and (A) and (B) of (iii), but not sure how to approach (C). Can someone please give some hints?
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To start, |(z-2)/(z+2)| = |z-2|/|z+2|.

Do you know how to find the modulus of z-2 and z+2? Nothing too fancy here, just trig.
(Note: Your diagrams for (A) and (B) are quite useful). Just add some measurements in and the approach should be obvious.)

[ 本帖最後由 風之男 於 2022-3-8 07:28 AM 編輯 ]

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回覆 3# 風之男 的帖子

Oh yes, thank you. I used cosine formula to find the modulus and solved it.

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Any suggestions for these two?

For Q5, I did the part (a), (b) and (c), but unsure how to prove the part (d).

For Q17, I think a possible way to do the question is to substitute all the values into the original equation and show it equals 0, but there should probably be a faster method.
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Hmm... I'm unsure what you've learnt and what you haven't learnt (again, it would be really helpful if you can at least show me what you've worked on so far, so I know what direction I'll try to explain without having to guess too much), so I'm just gonna throw out some questions.

5d) Do you know there is a formula for the product of 2 cosines? In fact, for part (c) there is also a formula for the sum of 2 cosines.
17) Have you learnt "roots of unity"?

[ 本帖最後由 風之男 於 2022-3-9 10:29 PM 編輯 ]

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For Question 5, this is where I got so far. I know there is formula for product of 2 cosines cosAcosB=(1/2)[cos(A-B)+cos(A+B)]. And for sum of 2 cosines, I know cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2], but not sure how it is applicable to the question.

For Question 17, I have learnt roots of unity, but still not sure how to approach the question.

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Sorry, I think the image didn't send

[ 本帖最後由 妙妙貓 於 2022-3-10 04:27 PM 編輯 ]
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5d)
So... did you actually use the formula? Because it's fairly straight-forward.
The odd part you might come across would be the 6pi/5, which... I'm gonna leave you to figure it out. It's not _too_ bad.

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17 is quite involved, that I'm tempted to bust out LateX.

But to use the idea of roots of unity, you would want to transform the equation into w^8=1, where w should be expressed in z.
Then... uh... you'll face with quite a bit of algebra. You might get stuck at anywhere. I'll let you try it first.

Again, because I don't know if you know Euler's formula, I'm gonna refrain from going that route (which, IMO, is _much_ less fuss).

Sidenote: googling this exact question (or rephrasing it slightly differently) will get you the solution.

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Oh yes, I got Q5. Originally, I didn't expand the cis into cos+isin and didn't notice that things would cancel.

And thank you for the hint for Q17, I solved it too.

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Could you please also give a hint about this one?

It defines w as a non-real cube root of unity, and I notice that z^2+z+1 is part of the factorisation of z^3-1=0 and if w is a non-real cube root of unity, then (substituting z as w), it will equal 0. But other than that, I have no idea about this question.
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引用:
原帖由 妙妙貓 於 2022-3-12 05:38 PM 發表
Could you please also give a hint about this one?

It defines w as a non-real cube root of unity, and I notice that z^2+z+1 is part of the factorisation of z^3-1=0 and if w is a non-real cube root  ...
P_n(x) has a factor of x^2+x+1 iff we can rewrite P_n(x) = (x^2+x+1)Q(x) for some polynomial Q(x).
Let's pick some special value of x, say w. What happens to the RHS of P_n(x)? Equals zero, right?

So if we can show that P_n(w) is indeed zero, then working backwards, substituting w back to x, the given claim is then proved.

Now your task is to show P_n(w)=0.

[ 本帖最後由 風之男 於 2022-3-12 06:51 PM 編輯 ]

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I proved that P_n(w)=0. But I'm a little confused as to why we can substitute the w back to x. In my working out, there is:

Because P_n(w)=0, (x-w) is a factor of P_n(x).

Because w^2+2+1=0 (x-w)=(x+1+w^2)

And then I can just substitute w as x to get "therefore (x^2+x+1) is a factor of P_n(x)"? Or did I misunderstand your meaning?

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引用:
原帖由 妙妙貓 於 2022-3-12 07:37 PM 發表
I proved that P_n(w)=0. But I'm a little confused as to why we can substitute the w back to x. In my working out, there is:

Because P_n(w)=0, (x-w) is a factor of P_n(x).

Because w^2+2+1=0 (x- ...
Ah, sorry for the faulty logic. You're right, it doesn't imply that.
I think I'd give you a not so obvious hint. You know x-w is a factor. Can you find _another_ factor such that when the two factors multiply, you get x^2+x+1?
Or in other words, can you find another value of x such that P_n(x)=0?
Usually with roots of unity questions, don't think too out of the picture. The other root is quite straight-forward.

If you're really stuck, I can give you the other factor as well.

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Oh thanks. I understand now. I used w^2 as the second factor and it worked.

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