# Proving that a three complex numbers form an equilateral triangle with vertices on unit circle [duplicate]

While solving questions from Mathematial Analysis by Apostol, I came across this question:-

Q. Given three complex numbers $z_1$, $z_2$ and $z_3$ such that $\left| z_1 \right| = \left| z_2 \right| = \left| z_3 \right| = 1$ and $z_1 + z_2 + z_3 = 0$. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with the center at origin.

Unfortunately, I could not get much ideas to solve this questions. However, from the given data, I inferred that the numbers are on the unit circle and hence, the vertices of the triangle are on the unit circle.

Also, I figured out that the centroid of this triangle is at origin. Now, if somehow we prove that the orthocenter of this triangle is also at origin, we will prove that the triangle is equilateral. But, I could not find a possible way to prove it by just using the information given in the question.

Another thought that I came across was to show that between any pair of the complex numbers (given) the difference in their argument, as measured from the positive X - axis, is $\dfrac{2\pi}{3}$ or $120^o$. If we prove this, we will prove that the triangle is equilateral. Again, there are no further ideas to implement this idea as well for the final proof!

Help will be appreciated!

## marked as duplicate by Community♦Jan 29 '18 at 11:55

• Show that $z_1+z_2+z_3 =0$ implies that the centroid and circumcenter, and hence orthocenter coincide. – Hari Shankar Jan 29 '18 at 11:34
Let us write $z_1,z_2,z_3$ in their polar representation: $$z_k=e^{i\theta_k},\ k=1,2,3$$ for some angles $\theta_k$. Now, since $$z_1+z_2+z_3=0\Leftrightarrow e^{i\theta_1}+e^{i\theta_2}+e^{i\theta_3}=0\Leftrightarrow1+e^{i(\theta_2-\theta_1)}+e^{i(\theta_3-\theta_1)}=0$$ or, $$e^{i(\theta_2-\theta_1)}+e^{i(\theta_3-\theta_1)}=-1\in\mathbb{R}$$ So, we get that $\Im(e^{i(\theta_2-\theta_1)}+e^{i\theta_3-\theta_1})=0$, or, using Euler's formula, that: $$\sin(\theta_2-\theta_1)+\sin(\theta_3-\theta_1)=0$$ Moreover, we have that $\Re(e^{i(\theta_2-\theta_1)}+e^{i(\theta_3-\theta_1)})=-1$, so: $$\cos(\theta_2-\theta_1)+\cos(\theta_3-\theta_1)=-1$$ For convenience, let $x=\theta_2-\theta_1$ and $y=\theta_3-\theta_1$, so we have: $$\left\{\begin{array}{l} \sin x+\sin y=0\\ \cos x+\cos y=-1 \end{array}\right\}$$ The first one, gives us that: $$\sin x=-\sin y=\sin(-y)\Leftrightarrow x=\left\{\begin{array}{l} 2k\pi-y\\ 2k\pi+\pi+y\end{array}\right.$$ Substituting in the second we take that either: $$2\cos y=-1\Leftrightarrow y=2k\pi\pm\frac{2\pi}{3}\text{ and }x=2kp\mp\frac{2\pi}{3}$$ or $$\cos(\pi+y)+\cos(y)=-1\Leftrightarrow-\cos y+\cos y=-1$$ which is not true.
So, $\theta_2,\theta_1$ and $\theta_3,\theta_1$ differ $\frac{2\pi}{3}$, which is equivalent to the triangle being equilateral.
Hint: If you divide the three points by $z_1$ you have three different points which satisfy the same conditions. One of these points is $1$ and that simplifies the analysis easily. Find those three points and then multiply by an arbitrary $z_1$ of modulus $1$ to get the general solution.