Looking for a good solution to system of complex equations

1st condition: The circumcenter of {z1, z2, z3} is {0,0}.
2nd condition: The centroid oif {z1, z2, z3} is also {0,0}.
Without going into the details, only when {z1, z2, z3} are vertices of an equilateral triangle both 1) and 2) are true. This is probably just saying that one can use complex numbers to prove certain things in geometry, and vice-versa. This is what makes math so fun.

Following @Micheal Mendrin, we just have to prove that $G\equiv O$ implies that $ABC$ is an equilateral triangle. This is a one-shot-one-kill proof through trilinear coordinates, or through the following observation - if $M$ is the midpoint of $AB$, then $G$ lies on $CM$ while $O$ lies on the perpendicular bisector through $M$, so $G\equiv O$ implies that $CM\perp AB$, or just $AC=AB$. In the same way we have $AB=BC$.

I saw this note right before I went to sleep last night and came up with two solutions. However, I see that @Michael Mendrin has beat me to it with the first one.

Here's the second one. We're going to use the fact that the resultant of two equal (in magnitude) vectors lies on the bisector of the angle between the two vectors. Its proof is pretty simple [and the fact is pretty intuitive as well].

Let the angle between $z_1$ and $z_2$ be $\alpha$. Let the angle between $z_2$ and $z_3$ be $\beta$ and the angle between $z_3$ and $z_1$ be $\gamma$.

Since, $z_1+z_2=-z_3$; $-z_3$ lies on the angle bisector of $\alpha$. So, $z_3$ also lies on the angle bisector of $\alpha$. That means $\beta=\gamma$. Similarly $\alpha=\beta$ and that means they are all equal to $120^\circ$.

I'm sorry if this is too short but it's almost midnight here. I'll add a diagram and do the vector proof tomorrow.

EDIT: Oh, by the way, you can not conclude that they are the vertices of an equilateral triangle. They don't form the vertices of an equilateral triangle when the numbers are all equal to zero :)

One solution that I have:

WLOG, we can rotate the system such that $z_1 = 1$. Let $z_2 = a + bi$, then the second condition gives us $z_3 = -1-a-bi$.

The first condition gives us $a^2 + b^2 = 1, (1+a)^2 + b^2 = 1$ and hence $a^2 = (1+a)^2 \Rightarrow a = - \frac{1}{2}$, which gives us that $z_1, z_2, z_3$ are the vertices of an equilateral triangle.

I would prefer a solution that isn't so algebraic / brute force.

I think treating complex numbers as vectors can help us

If we consider $z_{ 1 },z_{2},z_{3}$ as three vectors of equal magnitude,say $\boxed{k}$ (already given) then each of these vectors will be at 120° from each other to satisfy $z_{1}+z_{2}+z_{3}$ =0 Now we assume a circle of radius 'k' with center at the origin and obviously all three complex numbers will lie on the circle. We have seen that any two complex numbers subtend 120° at the origin which is indeed the center of circle. Now We can definitely say that the chord $|z_{2}-z_{1}|$ subtends 120° at the center so it'll subtend 60° on the circumference i.e.$z_{3}$ Similarly we can prove for the other two as well. Hence proved that it'll be an equilateral triangle

We can treat them as vectors and hence we get that they are making an angle 120○with each other ....or else we can consider the system as BF3 molecule where all bonds are equal and arranged in a way that they are as far away as possible....

If one triangle has the same centroid and circumcenter then it'e equilateral. It's easy to prove, because it means that every median is also perpendicular on each side.

Let the angle between $z_{1}$ and $z_{2}$ be $\alpha$, i.e. $z_{2} = z_{1} e^{i \alpha}$

Clearly, $|z_{1}+ z_{2}| = |-z_{3}|$

Square both sides to get:

$2 + 2 \cos \alpha = 1$

$\Rightarrow \cos \alpha = - \dfrac{1}{2}$

Hence, $\alpha = \dfrac{2 \pi}{3}$

Thus, $z_{2} = z_{1} e^{i \dfrac{2\pi}{3}} = z_{1} \omega$

$z_{3} = -(z_{1}+z_{2}) = z_{1} \omega^2$

Hence, the numbers are $z_{1} , z_{1} \omega , z_{1} \omega^2$, which are vertices of equilateral triangle, as the vectors from origin to these are rotated by $120^{\circ}$ to get the next vector.

From the second equation, by vector addition, the three values must be equally spaced apart (assuming this is referring to the Argand diagram). Thus the angle between each vector ($\theta$) must be equal to $120^\circ$, which is just $\frac{360}{3}$). From the first equation, their lengths are equal, thus they form the center of an equilateral triangle. A more rigorous proof:

Let $z_1=bi$ thus $z_1$ is pure imaginary, and when plotted is simply a vertical line lying on the imaginary axis. Without loss of generality, we will assume that the triangle is centered at the origin. Assuming the above is true ($\theta=120^\circ$), then $z_2$ and $z_3$ must be symmetric via. the complex axis with one in the third quadrant and one in the fourth. Thus, they form a $30^\circ$ angle with the real axis and their coordinates are $\ \pm \frac{\sqrt{3}|z_1|}{2}-|z_1|i/2$ by the laws of $30-60-90$ triangles. Now, let's add up all of their coordinates to see if it's zero. Obviously the $\pm$ causes the real portion to cancel out, leaving $2\frac{-|z_1|}{2}$ which is just $-z_1$. Adding that to the first leg ($z_1$) is zero and thus equation two is satisfied! Looking back at the coordinates for $z_2$ and $z_3$, we can use the Pythagorean theorem to verify that they have the same length as $z_1$. Plugging their coordinates in to the Distance Formula produces $(\frac{|z_1|^{2}}{4})+(\frac{3|z_1|^{2}}{4} \Longrightarrow (|z_1|)^{2}$ which shows that they all posses the same length and we're done! We've proved that the vectors form $120^\circ$ angles and all vectors have the same length, so the rest is just angle chasing! Letting $A$, $B$, and $C$ be the endpoints of each vector, and letting $O$ be the origin, we can see that $ABC$ is an equilateral triangle by connecting $AB$, $BC$, and $AC$. From there, we know that the angle $AOC$ is $120^\circ$, and it form an isosceles triangle ($OA$ and $OC$ are equal, remember?). Thus, the angles $OCA$ and $OAC$ are both $30^\circ$. This can be proven for the other two equal isosceles triangles which means that each angle $ABC$, $ACB$, and $BAC$ are $60^\circ$. And we're done! Yipee! *REMINDER TO ALL READERS, I AM A STUDENT IN ALGEBRA 1! :D