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Need help solving an quartic equation.

  1. Sep 28, 2011 #1
    Hi.

    1. The problem statement, all variables and given/known data

    Find the four roots of:
    https://www.physicsforums.com/attachment.php?attachmentid=39371&stc=1&d=1317233718

    2. Relevant equations
    Calculus: p.6 Polynomials and Rational functions

    3. The attempt at a solution
    Pretty much stuck on starting it. I can't really find a way to start factoring it.
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2011 #2
    Try using a substitution to turn it into a quadratic.
     
  4. Sep 28, 2011 #3
    hi spamiam, thanks for fast reply. Could you just show me fast how to get this one down to quadratic, only if you have time ofc. Have done similar equations with substitution, but this spesific one i can't get my head around.
     
  5. Sep 28, 2011 #4

    dynamicsolo

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    Concerning spamiam's suggestion, keep in mind that z4 can be written as ( z2 )2 ...
     
  6. Sep 28, 2011 #5

    SammyS

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    Let u = z2 or u = z2 - 1
     
  7. Sep 28, 2011 #6
    so;

    (z²)²-2z²+4=0 and let u=z²

    u²-2u+4=0

    u= 1+i√3 and u= 1-i√3

    And i find z

    z²= 1+i√3 and z²= 1-i√3

    z= √(1+i√3), z= -√(1+i√3), z= √(1-i√3), z=√(1-i√3)

    is this correct? i think the answer shud be more simplified
     
  8. Sep 28, 2011 #7

    dynamicsolo

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    If you write the result in Cartesian form (a + ib), you won't be able to "simplify" this much. Do you know how to write complex numbers in polar form? You could then get "simpler" expressions using DeMoivre's Theorem.
     
  9. Sep 28, 2011 #8
    dynamicsolo is right, polar form is much cleaner, especially if you note that [itex] \frac{1\pm i\sqrt{3}}{2}[/itex] are primitive sixth roots of unity and can be written [itex]e^{\frac{2 \pi i}{6}}[/itex] and [itex]e^{2 \pi i\frac{5}{6}}[/itex].
     
  10. Sep 28, 2011 #9
    Hi, dynamicsolo, yes shud know it, but just started my course so im abit new to polar form. I need to find modulus and argument of the complex number first right?
     
    Last edited: Sep 28, 2011
  11. Sep 28, 2011 #10
    Complex numbers can be depicted as points in the plane. Just draw yourself a triangle: the real axis is horizontal, and the imaginary is vertical. This gives you a triangle with what side lengths? Then the modulus is just the length of the hypotenuse, and the argument is the angle between the hypotenuse and the horizontal axis.
     
  12. Sep 28, 2011 #11

    epenguin

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    But for this problem #2 and #4 have as near told you what to do as anyone could without writing out your work for you.
     
  13. Oct 2, 2011 #12
    Still abit confused

    First root z= √(1+i√3)
    simplifed with polarform and DeMoivre's Theorem:

    modulus: √(1^2 + √3^2 ) = 2
    argument: tan = √3/1 = pi/3

    and polarform w=( 2*(cos(pi/3) + i sin(pi/3) )^1/2 ,and then with the theorem i can write: √2 (cos(1/2*pi/3) + i sin(1/2*pi/3))

    And then i get: √6/2 + i*√2/2 as my first root.

    but for next root z= -√(1+i√3) i have a problem. What do i do when i have -√?

    either i plug that in at the end and get same as first root just -(root 1)

    or i switch operator z= -√(1+i√3) -> z= √(1-i√3), but then i get same root as my 3th root z= √(1-i√3)


    Any tip what to do?:)
     
    Last edited: Oct 2, 2011
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