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Binomial formula (HELP)

  1. Nov 23, 2005 #1
    Hello

    I'm suppose to show

    Given z^4 + z^3 + z^2 + z + 1 = 0 where

    [tex]z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})[/tex]

    by using the binomial product formula.

    r^n - s^n

    Is that then

    if r,s = z then z^4 - z^4 = 0 ?????

    Sincerely and Best Regards

    Bob
     
  2. jcsd
  3. Nov 23, 2005 #2

    matt grime

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    i don't see any binomial product formula in your post, i just see r^n-s^n, surely that should equal something...?
     
  4. Nov 23, 2005 #3
    Theorem

    [tex]r^n - s^n = (r-s) \sum_{j=0} ^{n-1} r^{j} s^{n-j-1}[/tex]

    where r,s are numbers in the Ring R and n is a natural number.

    I'm presented with the following polymial

    z^4 + z^3 + z^2 + z +1 = 0

    where [tex]z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})[/tex]

    To show the above I'm told to use the theorem:

    Do this then imply that

    z^4 - z^4 = (z-1)(z^4 + z^3 + z^2 + z +1) = 0

    ???

    /Bob
     
  5. Nov 23, 2005 #4

    HallsofIvy

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    Yes, it is certainly the case that z^4- z^4= 0 no matter what z is!

    But, also certainly, that is not the question you intended to ask.

    It is true that r^n- s^n= (r- s)(r^(n-1)+ r^(n-2)s+ ...+ rs^(n-1)+ s^(n-1). (Although that is not waht I would call the "binomial product formula.)

    So if z^4 + z^3 + z^2 + z + 1 = 0, multiplying both sides by z-1,
    z^5- 1= 0 ro z^5= 1 and so z is a "root of unity". Was that what you intended?
     
  6. Nov 23, 2005 #5

    matt grime

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    good, now how can you get z^4+z^3+z^2+z+1 into a formula like that? setting r=z will help, now what about a choice for r?


    [/quote]where [tex]z = cos(\frac{2 \pi}{5}) + i sin(\frac{2 \pi}{5})[/tex]
    To show the above I'm told to use the theorem:
    Do this then imply that
    z^4 - z^4 = (z-1)(z^4 + z^3 + z^2 + z +1) = 0
    ???
    /Bob[/QUOTE]

    the fact that z^4=z^4 implies that z^4-z^4=0.

    your last line of maths is not an example of the binomial product formula
     
  7. Nov 23, 2005 #6
    Hello Hall,

    it is that thank Yoy :smile:

    But there is a second equation which has caused me some trouble too:

    z^8 + z ^ 6+ z^4 + z^2 +1 = 0

    z is the same as in the first one, but how I get i simular result as in the first one? If I multiply left and right by (z-1). I don't get an equation on the form z^n = 1.

    Any idears on how I solve this one?

    Sincerley
    Bob
     
  8. Nov 23, 2005 #7

    matt grime

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    all of the powers of z are even, aren't they,so it really is an equation in z^2,m isn't it?
     
  9. Nov 23, 2005 #8
    yes but if I divide the equation with z^2 I get

    z^4 + z ^ 3 + z ^2 + z + 1/z^2 = 0

    how do You get z^ 2 from that?

    Sincerely
    Bob
     
  10. Nov 23, 2005 #9

    matt grime

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    Firstly dividing by z^2 won't help, and secondly, even if you did divide by z^2 that isn't the result.
     
  11. Nov 23, 2005 #10
    Okay I misunderstood

    I have now multiplied it by (z^(2) -1) and got

    z^16 + z ^12 - 1 = 0 or z^16 + z^12 = 1

    Am I on the right track now ?

    Sincerley
    Bob
     
  12. Nov 23, 2005 #11

    matt grime

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    why are you making it more complicated? when i said it was a polynomial in z^2 what did I mean by that?

    Also, you're starting from the answer. you should not equate z^8+z^6+z^4+z^2+1 to zero to begin with since that is what you are trying to show.

    let's use some different letters.

    if i write

    a^4+a^3+a^2+a+1

    and

    b^8+b^6+b^4+b^2+1

    can you see a simple way to transform the second quantity into the first?
     
  13. Nov 23, 2005 #12
    You meant that z^2 is a root for the polynomial ?

    thereby (z^2)^4 + (z^2)^3 + (z^2)^2 +1 = 0

    /Bob

     
    Last edited: Nov 23, 2005
  14. Nov 23, 2005 #13

    matt grime

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    z^2 is not a root of the polynomial, though two things are being conflated there. but you managed to get the correct notion about making it a polynomial in z^2
     
  15. Nov 23, 2005 #14

    If I don't divide or multiply by z^2 and z^2 is not a root of the polynomial, how in God's name is it transformed into z^2 = 0 ??

    The only way I can see that is

    z^8 + z^6 + z^4 + z^2 +1 = z^8 + z^ 6 + z^4 + 1

    and then get z^2 = 0

    Sincerely

    Bob
     
    Last edited: Nov 23, 2005
  16. Nov 23, 2005 #15

    matt grime

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    who said it was transformed into that?

    you know that

    w^4+w^3+w^2+w+1= (w^5-1)/(w-1)

    that was the first thing you proved.

    and you know that w=z^2 gives what you want to examine. and you know that z is a fifth root of unity.

    incidentally, you do realize these are just geometric progressions?
     
    Last edited: Nov 23, 2005
  17. Nov 23, 2005 #16
    Hi Matt and thank You,

    Sorry that I have been slow to understand your surgestions.

    Following the first equation,

    then the second eqaution:

    z^8 + z^6 + z^4 + z^2 +1 = (z^10 + z^8 + z^6 + z^4 + z^2)/(z^2)

    Am I on the right path now?

    Sincerely and Best Regards,
    Bob

     
    Last edited: Nov 23, 2005
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