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Complex number method?

  1. Jun 19, 2013 #1
    complex number ??

    1. The problem statement, all variables and given/known data
    Let ω be the solution to the equation x2+x+1=0
    Get the value of ω105+3=


    2. Relevant equations
    complex numbers???


    3. The attempt at a solution
    When I try solving the first equation I hit a complex number which is making me think I am wrong.
    (x+1/2)2=-3/4
    Again if the method is right, what is the relationship between the complex number and the later expression?
     
  2. jcsd
  3. Jun 19, 2013 #2

    NascentOxygen

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    Staff: Mentor

    There are two solutions to the quadratic, both are complex numbers.
     
  4. Jun 19, 2013 #3
    Solve the equation and find the two complex solutions for x.

    Now, you know that these values are equal to ω, apply De Moivre's theorem for complex numbers to the new expression.
     
  5. Jun 19, 2013 #4

    HallsofIvy

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    You don't have to solve the equation- in fact, you don't have to use complex numbers at all.

    From [tex]x^2+ x+ 1= 0[/tex] we get [tex]x^2= -(x+ 1)[/tex]. [tex]x^{10}= (x^2)^5= -(x+ 1)^5= -x^5- 5x^4- 10x^3- 10x^2- 5x- 1[/tex] so that [tex]x^{10}+ x^5+ 3= -5x^4- 10x^3- 10x^2- 5x+ 2[/tex].

    Now, continue using [tex]x^2= -(x+ 1)[/tex] to keep reducing the exponents until you have reduce it to a quadratic.
     
    Last edited: Jun 19, 2013
  6. Jun 20, 2013 #5
    I tried doing that but I really do not know how to go about the -10x3The cubic power keeps resurfacing?
     
  7. Jun 20, 2013 #6

    NascentOxygen

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    Maybe try:

    x^2 = -(x+1)

    Therefore, x.x^2 = -x(x+1) = -x^2 - x

    Now, substitute for x^2 the expression on the top line, and you have the equivalent for x^3. :smile:
     
  8. Jun 20, 2013 #7
    Wow good insight now I have the quadratic equation and am stuck again.
     
  9. Jun 20, 2013 #8

    HallsofIvy

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    Yes, you can reduce it to a quadratic. Now compare it to [tex]x^2+ x+ 1[/tex] which you know is 0.
     
  10. Jun 20, 2013 #9

    D H

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    Determining the value of [itex]x^3[/itex] is critical. You know that [itex]x^2 = -(x+1)[/itex]. Multiply both sides by x and simplify the right hand side.

    Hall's approach is valid, but it's even easier if you use [itex]x^5 = x^3 x^2[/itex] and [itex]x^{10} = (x^3)^3 x[/itex].
     
  11. Jun 20, 2013 #10
    Nice I got it.Thanks
     
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