# Vertices of equilateral triangle inscribed in the unit circle

Prove that if $z_{1}+z_{2}+z_{3}=0$ and $|z_{1}|=|z_{2}|=|z_{3}|=1$ then the points $z_{1},z_{2},z_{3}$ are the vertices of an equilateral triangle inscribed in the unit circle $|z|=1$.

My idea was the following: since the sum of the numbers has to be $0$ and they have equal modulus, the interior angles between them should be equal to $120º$.

Later I could use the Inscribed Angles Theorem to prove that

$$\arg \left(\frac{z_{j}-z_{i}}{z_{j}-z_{k}}\right)=\frac{1}{2}\arg\left(\frac{z_{i}}{z_{k}}\right)$$

and since all $\arg(z_{i}/z_{k})=120º$, all angles would be equal. So the triangle formed would be equilateral.

I am wondering if this is correct and if would there be a more organized way to prove this.

-
I think the real meat of the exercise is how you conclude "the interior angles between them should be equal". Once you get there, each side of the triangle must have length $2\sin(60^\circ)$, so it is equilateral. –  Henning Makholm Sep 23 '12 at 11:47

$|z_1+z_2|^2=|z_3|^2\Rightarrow z_1\bar{z_2}+z_2\bar{z_1}=-1\Rightarrow|z_1|^2+|z_2|^2-z_1\bar{z_2}-z_2\bar{z_1}=|z_1-z_2|^2=3$

So $|z_1-z_2|=\sqrt{3}$ and similarly you have $|z_2-z_3|=|z_3-z_1|=\sqrt{3}.$

-
thankyou. Nice solution –  Mykolas Sep 23 '12 at 14:48

Multiplying through by $z_1^{-1}$ will rotate the numbers without changing any angles or distances, so you may assume $z_1=1$. Then $z_2$ and $z_3$ must have cancelling imaginary parts but can't be additive inverses; since they are on the unit circle, this forces them to be conjugate. So, they have the same real part, which must be $-1/2$ to cancel $z_1$. But you know where the numbers are on the unit circle that have real part $-1/2$.

-
Thankyou @Gerry Myerson, very easy and clear –  Mykolas Sep 23 '12 at 14:48

Consider the points $0$, $z_1$, $z_1+z_2$, and $z_1+z_2+z_3=0$ of the complex plane (the first and last are the same). By what is given, they form an equilateral triangle with side $1$, and one vertex at the origin. Then it is geometrically obvious that either $z_1+z_2=\exp(\frac{\pi\mathbf i}3)z_1$ or $z_1+z_2=\exp(\frac{5\pi\mathbf i}3)z_1$, and from this it follows that either $$z_2=\Bigl(\exp(\frac{\pi\mathbf i}3)-1\Bigr)z_1=\exp(\frac{2\pi\mathbf i}3)z_1 \quad\text{and} \quad z_3=\exp(\frac{4\pi\mathbf i}3)z_1,$$ or $$z_2=\exp(\frac{4\pi\mathbf i}3)z_1\quad\text{and} \quad z_3=\exp(\frac{2\pi\mathbf i}3)z_1.$$

-

You are given three vectors $z_i\in{\mathbb C}$ summing to $0$; therefore they can be regarded as the "directed sides" of a triangle. Since all three of them have length $1$ this triangle is equilateral. It follows that $z_{i+1}$ is obtained from $z_i$ by a $120^\circ$ rotation, or in terms of complex numbers: $z_{i+1}=e^{2\pi i/3} z_i$.

-