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Complex numbers on the unit circle

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##z_1,z_2,z_3## be three complex numbers that lie on the unit circle in the complex plane, and ##z_1+z_2+z_3=0##. Show that the numbers are located at the vertices of an equilateral triangle.

    2. Relevant equations


    3. The attempt at a solution
    As far as I understand, I need to show that ##|z_1-z_2|=|z_1-z_3|=|z_2-z_3|##. I have tried a lot of things but no where near a solution. I know that ##|z_i|=1## ,(i=1,2,3), since it's the distance of the number from the origin. Also tried to write ##z_i## in terms of the other two(it gave me that the distance from ##z_i## to ##(-z_j)## is 1) , but still nothing. I am missing something and not sure what. Please, don't give me the answer, just a direction of thinking.

    Thank you.
     
  2. jcsd
  3. May 10, 2016 #2

    fresh_42

    Staff: Mentor

    Have you thought about the geometric figure? What does it mean to add points on the unit circle?
    And it might become easier to assume, e.g. ##z_3 = 1##. It's certainly symmetric so this won't cost you generality.
     
  4. May 10, 2016 #3
    Consider points ##A(z_1)##, ##B(z_2)##, and ##C(z_3)##. Triangle ##ABC## is equilateral iff ##C## is deduced from ##B## by a rotation of center ##A## and angle ##\pi / 3##. How do you write such transformation ?
     
  5. May 10, 2016 #4
    Thank you both.

    I know that multiplication of two complex numbers is about scaling and rotating. since ##e^{ix}\cdot e^{iy}=e^{i(x+y)}##, so I guess I would write something similar to that. But I can't see more than that. I would like to get another hint.

    Why don't I lose the generality if I put ##z_3=1##?

     
  6. May 10, 2016 #5

    fresh_42

    Staff: Mentor

    Any solution with ##z_3=1## can be transformed by rotation in any other solution and vice versa.
    If there would be no solution for ##z_3 = 1## then there weren't any.
    I'm not sure whether it helps since I've not solved the problem, but at least it makes the angles involved familiar.
     
  7. May 10, 2016 #6

    Math_QED

    User Avatar
    Homework Helper

    I don't know whether this is useful but what do you know about the solutions of z1^3 = z2 when we draw them in the complex plane?
     
  8. May 10, 2016 #7
    Imagine that you had to prove the reciprocal, that is, if ##ABC## is equilateral, then ##z_1 + z_2 + z_3 = 0##. This can be solved by an elementary geometric argument : ##\angle AOB = \angle BOC = \angle COA = 2\pi / 3##. So ## z_2 = e^{2i\pi/3} z_1 ## and ## z_3 = e^{2i\pi/3} z_2 ##. The cubic roots of 1 sum to zero so ##z_1 + z_2 + z_3 = 0 ##.

    Conversely, if ##z_1 + z_2 + z_3 = 0##, you must show that ## z_3 - z_1 = e^{i\pi/3} (z_2 - z_1) ##
     
  9. May 10, 2016 #8
    The problem statement states that all three points belong to the unit circle, so ##z_2## and ##z_3## can be deduced from ##z_1## by a rotation centered at the origin. That is to say there exists angles ##\theta## and ##\rho## to determine such that ##z_2 = e^{i\theta} z_1 ## and ## z_3 = e^{i\rho} z_1 ##.
    From the hypothesis, one deduces that ## 1 + e^{i\theta} + e^{i\rho} = 0##, which leads to a system of two equations : ## 1 + \cos \theta + \cos \rho = 0 ## and ##\sin\theta + \sin\rho = 0##.
    The second equation implies that ##\theta = - \rho ## or ## \theta = \pi + \rho ##, but the last option is incompatible with the first equation. So ##\theta = -\rho ## and ## \cos\theta = -1/2 \Rightarrow \theta \in \{ -2\pi / 3 , 2 \pi / 3 \} \text{ mod } 2\pi ##.
    This implies that the triangle is equilateral but it's up to you to verify this.
     
  10. May 11, 2016 #9

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Polya problem solving strategy: work on the simplest case of your problem.

    I would take the case where one of your z is as simple as possible. You can only choose one in this way because once chosen the others are determined. But then you can find out what these are. Then rotate the points you have determined through an arbitrary angle.

    If this is successful, then ask yourself what was stopping you from seeing it in the first place. Unless this is nearly the last maths course you will ever do, get hold of the book by Polya "How to Solve It".
     
  11. May 11, 2016 #10
    Hi,

    Thank you all for the answers. I have succeeded to show that if z1=1 then ##z2=e^{i\frac{2pi}{3}}## and ##z3=e^{i\frac{4pi}{3}}## but can't understand how to prove that it implies that it must be an equilateral triangle, not using geometry(it is easy to show it geometrically, but since this is a course in algebra, I don't use geometry here). I have tried to calculate and found that ##|z1-z3|=|z1-z2|=\sqrt{3}## but failed for ##|z2-z3|##. Can you help me here?

    Thank you all!
     
  12. May 11, 2016 #11

    fresh_42

    Staff: Mentor

    In this case just write ##z_2 = -\frac{1}{2}+i\sqrt{\frac{3}{4}}## and ##z_3 = -\frac{1}{2}-i\sqrt{\frac{3}{4}}##. But I don't see a way to avoid trigonometric functions.
     
  13. May 11, 2016 #12
    Ok, apparently it was a mistake of signs. I have got all of them sqrt(3). Problem solved. Thank you all!
     
  14. May 9, 2017 #13
  15. May 9, 2017 #14
    Can someone hit me up with the full solution? Really interesting! Im not at this level yet but it would be really cool to see the proof!
     
  16. May 9, 2017 #15

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    We are not allowed to do that---PF rules, I'm afraid.
     
  17. May 9, 2017 #16
    Ah okey but somewhat of a solution then? Ish?
     
  18. May 10, 2017 #17

    StoneTemplePython

    User Avatar
    Gold Member

    An interesting way to encode the original poster's problem / constraints would be via the following 3x3 permutation matrix

    ##\mathbf P =\begin{bmatrix}
    0 & 0 & 1\\
    1 & 0 & 0\\
    0 & 1 & 0
    \end{bmatrix}##

    Notice that the eigenvalues must sum to zero via trace. Since the matrix is orthogonal, we know that all eigenvalues have magnitude of 1. This gives all the information needed to prove (or at least interpret) the original claim in the case of ##\alpha \mathbf P## where ##\alpha## itself is some complex scalar on the unit circle.

    (As a bonus, the above structure tells us things like ##\mathbf P \mathbf 1 = \mathbf 1##, because all permutation matrices are (doubly) stochastic, and that any complex eigenvalues for ##\mathbf P## come in conjugate pairs. Also note that I could have used a different 3x3 Permutation matrix than the above -- of the 6 possible 3x3 Permutation matrices I could use any of the three that do not have zeros on the diagonals (to observe the constraint that the trace = 0).)
     
    Last edited: May 10, 2017
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