Complex numbers on the unit circle

[mr.tea]

Homework Statement

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.

Homework Equations

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.

 

[fresh_42]

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.

 

[geoffrey159]

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 ?

 

[mr.tea]

Thank you both.

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 ?

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.

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.

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

 

[fresh_42]

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

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.

 

[member 587159]

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?

 

[geoffrey159]

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 $z_3 - z_1 = e^{i\pi/3} (z_2 - z_1)$

 

[geoffrey159]

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.

 

[epenguin]

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".

 

[mr.tea]

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!

 

[fresh_42]

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!

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.

 

[mr.tea]

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.

Ok, apparently it was a mistake of signs. I have got all of them sqrt(3). Problem solved. Thank you all!

 

[Lotta]

..

 

[Lotta]

Can someone hit me up with the full solution? Really interesting! I am not at this level yet but it would be really cool to see the proof!

 

[Ray Vickson]

Can someone hit me up with the full solution? Really interesting! I am not at this level yet but it would be really cool to see the proof!

We are not allowed to do that---PF rules, I'm afraid.

 

[Lotta]

Ah okey but somewhat of a solution then? Ish?

 

[StoneTemplePython]

Ah okey but somewhat of a solution then? Ish?

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).)