Vertices of equilateral triangle inscribed in the unit circle

Solution 1:

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 $-1/2$.

Solution 2:

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

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

Solution 3:

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: $z_{i+1}=e^{2\pi i/3} z_i$.

Solution 4:

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 $z_1+z_2=\exp\left(\frac{\pi\mathbf i}3\right)z_1$ or $z_1+z_2=\exp\left(\frac{5\pi\mathbf i}3\right)z_1$, and from this it follows that either $$ z_2=\Bigl(\exp\left(\frac{\pi\mathbf i}3\right)-1\Bigr)z_1=\exp\left(\frac{2\pi\mathbf i}3\right)z_1 \quad\text{and} \quad z_3=\exp\left(\frac{4\pi\mathbf i}3\right)z_1, $$ or $$ z_2=\exp\left(\frac{4\pi\mathbf i}3\right)z_1\quad\text{and} \quad z_3=\exp\left(\frac{2\pi\mathbf i}3\right)z_1. $$