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This question already has an answer here:

Suppose that $z_1+z_2+z_3=0$ and $|z_1|=|z_2|=|z_3|=1$. I want to prove that $z_1,z_2,z_3$ are the vertices of an equilateral triangle.

We can assume that $z_1\neq0$. Thus $1+\frac{z_2}{z_1}+\frac{z_3}{z_1}=0$ and $|\frac{z_2}{z_1}|=|\frac{z_3}{z_1}|=1$.

I can prove that if $1+z_2+z_3=0$ and $|z_2|=|z_3|=1$, then $1,z_2,z_3$ are vertices of an equilateral triangle.

So back to original problem we can conclude that $1,\frac{z_2}{z_1},\frac{z_3}{z_1}$ are vertices of an equilateral triangle. Could you give me some hint to continue from here? Thanks!

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marked as duplicate by Martin R, Did, Robert Soupe, Namaste algebra-precalculus May 8 '18 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Update. First I proposed the following hint: Think of three unit vectors being added geometrically. But it was rightfully commented that you expected an algebraic solution. Therefore I'm starting afresh.

Note that your sentence "I can prove that if $1+z_2+z_3=0$ and $|z_2|=|z_3|=1$, then $1$, $z_2$, $z_3$ are vertices of an equilateral triangle" immediately leads to a proof since after a rotation you can always assume that $z_1=1$.

Nevertheless I'd remark that your approach destroys the inherent symmetries of the problem. Before making such a move you should try to make full use of these symmetries. Here is a proposal:

From $$0=\bar z_1+\bar z_2+\bar z_3={1\over z_1}+{1\over z_2}+{1\over z_3}={z_2z_3+z_3z_1+z_1z_2\over z_1z_2z_3}$$ it follows that $z_2z_3+z_3z_1+z_1z_2=0$ as well. By Vieta's theorem we therefore can conclude that the three $z_i$ are the solutions of an equation of the form $$z^3+a=0\tag{1}$$ with $a=-z_1z_2z_3\ne 0$. Now if $z\in{\mathbb C}$ is a solution of $(1)$ then the numbers $\omega z$, $\omega^2 z$, where $\omega:=e^{2\pi i/3}$ is a complex third root of unity, are solutions as well. It follows that the numbers $z_i$ form an equilateral triangle with vertices on the unit circle.

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  • $\begingroup$ I think OP wants an algebraic solution. Since putting complex variables into vectors trivialize this problem... $\endgroup$ – Macrophage May 8 '18 at 7:36
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Could you give me some hint to continue from here?

Multiplying each of $1,\frac{z_2}{z_1},\frac{z_3}{z_1}$ by $z_1$ is just a rotation. In fact, multiplying by any complex number is just a rotation and a scaling. Thus it can't ruin the equilateral-triangle-ness of your three points.

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Let be $z_k = x_k + y_ki$. As you said, we can suppose wlog $z_1 = 1$. Then, $$ 1 + z_2 + z_3 = 0\implies x_2 + x_3 = -1,\ y_2 + y_3 = 0. $$ Now, $$ x_2^2 + y_2^2 = |z_2|^2 = 1 = |z_3|^2 = x_3^2 + y_3^2\implies |x_2| = |x_3|\implies x_2 = x_3 = -\frac12\implies\cdots $$

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If you are allowed to use that the 3rd roots of $1$ form an equilateral triangle, you can do the following:

  • WLOG: $z_1 = 1$ (otherwise rotate the whole thing by $e^{-arg(z_1)}$)
  • $1+z_2+z_3=0 \stackrel{z_2+z_3=-1, |z_2|=|z_3|=1}{\Rightarrow} $$z_2= z_0$, $z_3 = \overline{z_0}$ for some $|z_0|=1,\; z_0 \neq 1$
  • $\Rightarrow 1,z_0,\overline{z_0}$ satisfy $(z-1)(z^2+z+1)= z^3-1=0$.
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$z_1 + z_2 = -z_3 $ and hence $\bar{z_1} + \bar{z_2} = -\bar{z_3}$. Multiplying, $2 + 2(z_1\bar{z_2} + z_2\bar{z_1}) = 1$ and this gives that angle between the lines joining the origin to $z_1$ and $z_2$ is $2\pi/3$. Since the circum center of the triangle is the origin, the vertex angle is $\pi/3$ and the triangle is equilateral.

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Note that $z_1=-z_2-z_3$. Hence, since $|z_1|=|z_2|=|z_3|=1$, $$|z_2-z_1|^2=(z_2-z_1)(\bar{z_2}-\bar{z_1})=2-z_1\bar{z_2}-z_2\bar{z_1}= 4+z_3\bar{z_2}+z_2\bar{z_3}$$ and $$|z_3-z_1|^2=(z_3-z_1)(\bar{z_3}-\bar{z_1})=2-z_1\bar{z_3}-z_3\bar{z_1}= 4+z_2\bar{z_3}+z_3\bar{z_2}.$$ Therefore $|z_2-z_1|=|z_3-z_1|$.

By symmetry we may conclude that $|z_2-z_1|=|z_3-z_1|=|z_3-z_2|$, that is the triangle is equilateral.

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