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!

Complex numbers

  1. Jan 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that ##(1+u^2)^2+(1+u)^2>0##
    where ##u=exp(\frac{2\pi i}{3})##


    2. Relevant equations



    3. The attempt at a solution
    ##u^4=u## so
    ##(1+u^2)^2+(1+u)^2=3u^2+3u+2>0##
    is there any way from here to get ##(1+u^2)^2+(1+u)^2>0##?
     
  2. jcsd
  3. Jan 24, 2014 #2
    Note that [itex]u^2=u^*[/itex] and so [itex]u^2+u[/itex] is real
     
  4. Jan 24, 2014 #3
    Maybe this ##3u^2+3u+2>0## because discriminant is complex? Another way?
     
  5. Jan 24, 2014 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi LagrangeEuler! :smile:

    Hint: u3 - 1 = 0 and u - 1 ≠ 0, sooo … ? :wink:

    (hmm … is it true?)
     
  6. Jan 24, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In fact, do you have any reason to think that [itex](1+ u^2)^2+ (1+ u)^2[/itex] is even a real number?

    Here is my very simple-minded, not at all sophisticated look at it:

    [tex]u= e^{2\pi i/3}= -1- i\sqrt{3}/2[/tex]
    [tex]u^2= \frac{1}{4}+ i\sqrt{3}[/tex]
    [tex]1+ u^2= \frac{5}{4}+ i\sqrt{3}[/tex]
    [tex](1+ u^2)^2= -\frac{13}{16}+ 5i\sqrt{3}/2[/tex]
    While [itex]1+ u[/itex] is [itex]-i\sqrt{3}/2[/itex] so [itex](1+ u)^2[/itex] is real and the sum has non-zero imaginary part.
     
  7. Jan 24, 2014 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have ##3u^2+3u+2## and it has been noted that ##u^2 = \bar u## so you have ##3u+3\bar u + 2 = 3(u + \bar u)+2 = 3\cdot 2\mathcal{Re} (u) + 2##, which is easy enough to calculate.
     
    Last edited: Jan 24, 2014
  8. Jan 24, 2014 #7
    This is incorrect since [itex]|u|=1[/itex] it should be [itex]u= -1/2- i\sqrt{3}/2[/itex] from which the algebra follows
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complex numbers
  1. Complex Numbers (Replies: 5)

  2. Complex number method? (Replies: 9)

  3. Complex Numbers (Replies: 2)

  4. Complex numbers (Replies: 6)

Loading...