1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Different complex numbers

  1. Jan 4, 2013 #1
    We have a,b,c different complex numbers so

    (a+b)^3 = (b+c)^3 = (c+a)^3

    Show that a^3 = b^3 = c^3

    From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0 How a is different from c => a-c is different from 0
    How do I show that a^3 - c^3 = 0?
     
  2. jcsd
  3. Jan 4, 2013 #2
    If you can prove that 3b(a-c)(a+b+c)=0, you're done.
    a, b, c are complex numbers.
     
  4. Jan 4, 2013 #3
    But how to prove that?
     
  5. Jan 4, 2013 #4
    Well, you can try assigning a value to the complex numbers and put them in the expression.
     
  6. Jan 4, 2013 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Huh? How can that ever be a good proof??
     
  7. Jan 4, 2013 #6
    lol i think it can't be..but you can do it if you take the time..
    edit: expand the expression first.
     
    Last edited: Jan 4, 2013
  8. Jan 4, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's a good start. (It will turn out that a+b+c=0.) If you take out the factor a-c, and also write down a corresponding equation with a/b/c rotated around, then add the two, what do you get?
     
  9. Jan 4, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Doh! There's a much easier way.
    We know (a+b) = (b+c)ω, where ω3=1. If a≠c then ω≠1. Writing the corresponding eqn for c+a versus b+c, and assuming b≠a and b≠c, we have (b+c) = (c+a)ω. (If (b+c) = (c+a)ω2 then b=c.) It's not hard from there.
     
  10. Jan 5, 2013 #9
    redount2k9, this formula can help you:

    ##x^3-y^3=(x-y)(x^2+xy+y^2)##

    You will need to prove 3b(a-c)(a+b+c)=0 using this forumla, as I said earlier. Or you can put ##a=x_1+iy_1, b=x_2+iy_2, c=x_3+iy_3##, but this will become lengthy.
     
  11. Jan 5, 2013 #10

    Curious3141

    User Avatar
    Homework Helper

     
  12. Jan 5, 2013 #11
     
  13. Jan 5, 2013 #12
    I don't know if the original poster is still with us...

    We know a, b, and c are distinct.

    Therefore (a+b), (b+c) and (c+a) are also distinct

    Therefore , as per the question, they are the three distinct cube roots of some number z.

    And therefore what is the sum ((a+b) + (b+c) + (c+a))?
     
  14. Jan 5, 2013 #13
    2(a+b+c)
     
  15. Jan 5, 2013 #14
    More specifically, what is the sum of the three distinct cube roots of some complex number z?
     
  16. Jan 5, 2013 #15
    That's 0 ..there you go. This can be locked. :P
     
  17. Jan 5, 2013 #16

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Anyway. Use that if

    [tex](a+b)^3=(b+c)^3[/tex]

    then

    [tex]a+b=e^{2ik\pi/3}(b+c)[/tex]

    for [itex]k\in \mathbb{Z}[/itex]. And the same thing for the equality [itex](a+b)^3=(a+c)^3[/itex].

    Now solve the equations

    [tex]\left\{\begin{array}{l} a+b=e^{2ik\pi/3}(b+c)\\ a+b=e^{2ik^\prime\pi/3}(a+c) \end{array}\right.[/tex]
     
  18. Jan 5, 2013 #17

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Thread is open again. Let's actually wait for the OP to reply before helping any further :tongue2:
     
  19. Jan 6, 2013 #18
    What should I reply? All of you have different opinions... it's hard for me to understand something.
     
  20. Jan 6, 2013 #19
    hahahaha use any of the methods..
     
  21. Jan 6, 2013 #20
    Hahahaha why not to use the best method? (if I would know which it is)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Different complex numbers
  1. Complex number method? (Replies: 9)

  2. Complex Numbers (Replies: 2)

  3. Complex numbers (Replies: 6)

  4. Complex numbers (Replies: 6)

Loading...