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Need help with roots of unity. (1 Viewer)

Run hard@thehsc

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How would you do part 5 of this question? I have done questions like these before, but somehow can't find the method I used before. Thanks!
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View attachment 34434
I did part 2 but I need help to confirm weather my method was correct... can someone show how this problem can be approached??
has roots

let be w, be w conjugate and j for the other root etc.


dividing z^5 -1 with z -1 gets the RHS






same with the j root.

and then you get the result as required.

edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
edit 2: fixed a typo
 
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Vipul_K

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has roots

let be w, be w conjugate and j for the other root etc.


dividing z^5 -1 with z -1 gets the RHS






same with the j root.

and then you get the result as required.

edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
I understand this but can u please expand on how dividing z^5-1 with z-1 gets the rhs? Thanks!
 
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I understand this but can u please expand on how dividing z^5-1 with z-1 gets the rhs? Thanks!
You can prove it through polynomial division. If you were to manually divide both polynomials, you would get that term.

Or

You could factorise z^5 - 1 into two polynomials



as you can see, all the terms in the polynomial are multiplied by z first, and then the regular polynomial is multiplied by -1 so all the terms between z^5 and -1 would be cancelled out


which is


so when you divide by z - 1.



*z can not equal to 1 btw

edit: i forgot the - z
 
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Vipul_K

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You can prove it through polynomial division. If you were to manually divide both polynomials, you would get that term.

Or

You could factorise z^5 - 1 into two polynomials



as you can see, all the terms in the polynomial are multiplied by z first, and then the regular polynomial is multiplied by -1 so all the terms between z^5 and -1 would be cancelled out


which is


so when you divide by z - 1.

Ahhhhh okkk makes sense now thank you!!
 

CM_Tutor

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edit: don't know how to do conjugates so if anyone reading could show how to latex it i would appreciate
You can use \bar{z} to get but I prefer \overline{} so I can get \overline{z} is and \overline{a+ib} is because \bar{a+ib}, giving looks ridiculous.
 
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How would you do part 5 of this question? I have done questions like these before, but somehow can't find the method I used before. Thanks!
View attachment 34431
Not the solution to (v) but just another method for these type of questions;

Here's what I did.







Sub this in




from (iv), the exact value is for is

Therefore;



as required.
 
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Lith_30

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I think this might be the solution to part v

We know that



We can do a similar thing with but we sub in

and the result should be after equating the real components...



therefore by multiplying the two results we get

 

CM_Tutor

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If part (iv) has established that


then applying the result


shows that


from which we can deduce that


Now, we seek

 

Lith_30

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I understand... I was wondering whether it meant to use part (iv), or about noting that part(iv) was derived from (ii), and so using it is building from part (ii)
I think the question would have explicitly stated to use part (iv) if they meant for you to use it. But it I guess the question was kinda building up to that point through all the previous parts, so your way would have been correct too and in timed conditions my method would take way to long for 2 marks.
 

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