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Complex Numbers using Argand Diagram

  1. Jun 17, 2009 #1
    1. The problem statement, all variables and given/known data
    (z^3)-1 = 0. Solve the equation and show that the roots are represented in an Argand Diagram by the vertices of an equilateral triangle.

    2. Relevant equations

    3. The attempt at a solution
    i can only found 1 as the answer.By transferring the -1 to the other side to become 1 and cube root it.

    The prob is, is there another answer to this question?
  2. jcsd
  3. Jun 17, 2009 #2


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    Hi cyy91! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    The clue is in the word "diagram"!

    Draw the Argand plane, and remember every number is of the form re

    what circle or triangle are the answers on? :wink:
  4. Jun 18, 2009 #3
    the answers indicate
    z = 1
    z = -[tex]\frac{1}{2}[/tex] +/- [tex]\frac{square root 3}{2}[/tex] i

    i understand dat when a triangle is equilateral, all its angle must be 60 degrees.
    with that in mind i can use cos 60 to find x and sin 60 for y...
    d oni thg i dun understand nw is da positive n negative signs...
    my answer is in positive for both instead of wad da answer indicate...
  5. Jun 18, 2009 #4


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    You clearly are expect to know DeMoivre's theorem: that if [itex]z= x+ iy= rcos(\theta)+ rsin(\theta)i= r(cos(\theta)+ i sin(\theta))[/itex], then [itex]z^n= r^n(cos(n\theta)+ i sin(n\theta))[/itex]. That works with fractional n also. If n= 1/m, then [itex]z^n= z^{1/m}= r^{1/m}(cos((\theta+2k\pi)/m)+ i sin((\theta+ 2k\pi)m)[/itex].

    I have no idea what you are trying to say in your last post.
    Last edited by a moderator: Jun 19, 2009
  6. Jun 18, 2009 #5


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    Yeah that last post was an eye-sore. Save the msn talk for txt msgs :tongue:

    More simply, try factorising.
    Using the difference of two cubes:


    So, [tex]z^3-1=(z-1)(z^2+z+1)=0[/tex]

    Now you have two factors. A linear factor that gives z=1, and a quadratic that has unreal solutions. Solving this quadratic for z will give you the two imaginary roots of [itex]z^3=1[/itex] and if you know how to plot complex numbers on an argand diagram, then it should be easy from there.

    p.s. all 3 solutions will lie on a circle of unit radius on the argand diagram.

    EDIT: didn't read the whole question.

    ok so to show the points lie on an equilateral triangle, are you aware that if you multiply a vector on the argand diagram by [itex]cis\theta[/itex], it will rotate the vector [itex]\theta[/itex] in the anti-clockwise direction?

    If you don't understand what I just said, then just show that the distance between all the solutions are equal, thus creating an equilateral triangle.
    Last edited: Jun 18, 2009
  7. Jun 19, 2009 #6
    thx a lot metallic...
    nw i understand...
    cuz my teacher nw oni say dat dis question has haven been taught...
    it requires long division which u jz xplained...thx
  8. Jun 19, 2009 #7


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    No problem :smile:

    Uh just curious, but the time you possibly save cutting down half of the true spelling of each word couldn't be worth the time the reader wastes trying to follow what you intended to say.
    But I'm taking into consideration that you're probably msn addicted and unable to touch-type yet. Just wait for it - one day it will strike you that spelling a word correctly and structuring it in a valid sentence is much more rewarding for both typer and reader. One day :biggrin:
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