妙妙貓

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?

風之男

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 ]

妙妙貓

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

妙妙貓

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.

風之男

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 ]

妙妙貓

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.

妙妙貓

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 ]

風之男

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.

風之男

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.

妙妙貓

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.

妙妙貓

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.

風之男

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 ]

妙妙貓

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?

風之男

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.

妙妙貓

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