Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex polynomial roots?

  1. Apr 14, 2005 #1
    Hi guys, can anyone tell me how I would go about solving this equation? :

    [itex]x^5 = x[/itex]

    Rearranging it gives:
    [itex]x^5 - x = 0[/itex]

    But then I don't really know what to do next. I know just from looking at it and thinking about it that the roots should be x = 0, 1, -1, -i, i...but I need to be able to come to that conclusion rather than state it.

    Thanks for any help :)
     
  2. jcsd
  3. Apr 14, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Well, you can clearly see x=0 is a solution, so for the other solutions, you can divide the equation by x.
    Can you solve the remaining equation? (There's a general method for finding the n n-th roots of unity).

    By the way, since you have all the solutions, you can factor the polynomial or so.
     
    Last edited: Apr 14, 2005
  4. Apr 14, 2005 #3
    ahh of course! That'll explain it. I did wonder as i was looking through my notes, we were taught how to find roots of unity but not this. Dividing by x makes that possible, cheers :)
     
  5. Apr 21, 2005 #4
    hey
    x can also be infinity
     
  6. Apr 21, 2005 #5
    you can simply factor it like this:

    [tex]x^5 - x = 0[/tex]
    [tex]x(x^4-1)=0[/tex]
    [tex]x(x^2+1)(x^2-1)=0[/tex]
    [tex]x(x^2+1)(x+1)(x-1)=0[/tex]

    now its obvious the real solutions are:
    [tex]0,-1,1[/tex]

    and the complex solutions are:
    [tex]x^2+1=0[/tex]
    [tex]x^2=-1[/tex]
    [tex]x=i[/tex]
    [tex]x=-i[/tex]
     
    Last edited: Apr 21, 2005
  7. Apr 21, 2005 #6
    The only way this is posible is if [tex]x:=0; 0^5 = 0;[/tex]
     
  8. Apr 21, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Of course,the complex exponential is multivalued,that means that the # of solutions to

    [tex] x^{4}=1 [/tex]

    is infinite.

    Daniel.
     
  9. Apr 21, 2005 #8
    nice work anzas,
    and u exulus,my idea was this divide both side by x,u get x^4=1call this equation one,and rem. x[x^4-1]=0 this was resolved from the above question sub. x^4as 1 x(0)=0now divide both sides by 0 x=0/0 which is infinity.if u disagree let me know .see ya
     
  10. Apr 23, 2005 #9
    bad idea. What does it mean to divide by zero exactly? And why do you think 0/0 is infinity?

    I always thought the complex exponential was pretty single-valued. What's a value of [itex]e^{ i\alpha}[/itex], other than [itex]\cos \alpha + i\sin \alpha[/itex]?
     
  11. Apr 24, 2005 #10

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    the complex exponential is single valued. maybe he meant the complex log.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook