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Simplification problem.

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    http://postimg.org/image/tgx9b3n3t/ [Broken]

    i drew a picture in paint of the initial question. Now me and my freind have both solved this question using different methods and have resulted in the same answer. The answer in the back of the book however is different from both of our answers. I'm wondering if some one can help me out and tell me if i'm missing something, or if the book is indeed wrong.


    2. Relevant equations

    (z^2 - z +1) / ((z^2 + 1/z^2)^2 + 2(z+1/z)^2 -3))^1/2



    3. The attempt at a solution

    Now what i did is i took the bottom piece and i subtracted and added 2 to make the first piece (z^2 + 1/z^2)^2 similar to the second piece

    ((z^2 + 1/z^2 +2 -2)^2 +2(z+1/z)^2 -3)^1/2

    =(( (z+1/z)^2 - 2)^2 +2(z+1/z)^2 - 3)^1/2

    when i simplify this i get

    [ ((z+1/z)^2 - 1)^2 ] ^1/2

    = z^2 - 2 + 1/z^2 -1

    = (z^4 + z^2 + 1)/z^2

    when i divide this peice by the initial numerator z^2 - z + 1

    my answer yields


    z^2(z^2 -z +1)/ (z^4 + z^2 -1)


    The back of the books answer is

    z^2/(z^2 + z +1)


    The only thing i could think of from my answer was that i could factor the bottom piece to cancel somethign with the top but the denominator is unfactorable from what i can see and what wolfram alpha can as well. sooo yea.. that's why i get stuck with my answer.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 3, 2013 #2

    DrClaude

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    Compare the first line and the last in the above quote.
     
  4. Apr 3, 2013 #3
    sorry i inverted that sentence.

    It should say if i take the numerator divided by that peice.

    ahh and it should say

    z^2(z^2-z+1)/(z^4 + z^2 +1)
     
    Last edited: Apr 3, 2013
  5. Apr 3, 2013 #4

    haruspex

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    ... And can you see how to simplify that? Tip: to see whether polynomial P is a factor of Q, set P=0 and use that eqn to reduce Q progressively until it's of order less than that of P. P is a factor if and only if the result is 0.
     
  6. Apr 4, 2013 #5
    so I didn't use your hint because I dont believe I am understanding it correct, but what I saw was that if I take z^4 + z^2 +1 and I say z^4 +z^2 + 1 + z^2 - z^2 it will become
    z^4 + 2z^2 + 1 - z^2
    =(z^2 + 1)^2 - z^2

    And this difference of squares will give you the factor. Alternatively my friend found you can multiply by (z+1)/(z+1) and this will also provide factors which will yield the answer. However I was curious if you could show me this factor method where 0=polynomial and you reduce the variable z in this case.
     
  7. Apr 4, 2013 #6

    SammyS

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    You appear to be saying that [itex]\displaystyle \ \left( z^2+\frac{1}{z^2}\right)^2=\left(\left(z+\frac{1}{z}\right)^2-2\right)^2\ ,\ [/itex] which is correct.

    Wow! That's easier to read in LaTeX !
    Look at

    [itex]\displaystyle \frac{z^2}{\displaystyle\left(\frac{z^4+z^2+1}{z^2-z+1}\right)}[/itex]

    using long division to simplify the denominator.
     
    Last edited by a moderator: May 6, 2017
  8. Apr 4, 2013 #7

    haruspex

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    Q(z) = z4+z2+1; P(z)=z2-z+1
    Setting P(z)=0 we have z2 = z-1.
    Substituting in Q:
    Q(z) = z2(z2+1)+1 = (z-1)(z-1+1)+1 = z2-z+1 = P(z) = 0. Therefore P(z) divides Q(z).
     
  9. Apr 4, 2013 #8
    Thank you sir.
     
    Last edited: Apr 4, 2013
  10. Apr 4, 2013 #9
    Okay so the method below to show that i can divide the function by the other function, and then using long division like you said. Thank you for your help
     
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