Equilateral triangle in complex number

Question

Let $a$, $b$ and $c$ be the affix of $A$, $B$ and $C$, where $a+b+c=0$ and $a$, $b$ and $c$ are of equal magnitude. Prove that $\triangle ABC$ is an equilateral triangle.

Answer 1

Let us take $a$ to be $1 + 0i$. (We can rotate the three vectors through the same angle, and this preserves the angles between them.) We then have $$1 + b + c = 0,$$ so $\Re b + \Re c = -1$ and $\Im b + \Im c = 0$. Solving for $b$ and $c$ gives $-\dfrac12 \pm \dfrac{\sqrt3}{2}i$.

Answer 2

Suppose without loss of generality that $|a|,|b|,|c| = 1$. This implies that \begin{align*} a &= e^{i\theta_1}\\ b &= e^{i\theta_2}\\ c &= e^{i\theta_3}\\ \end{align*} since $a + b + c = 0$ this means $e^{i\theta_1} + e^{i\theta_2} + e^{i\theta_3} = 0$ or, dividing by $e^{i\theta_1}$ that $e^{i(\theta_2-\theta_1)} + e^{i(\theta_3-\theta_1)} = -1 + 0i$. Continuing, $sin(\theta_2 - \theta_1) = sin(\theta_1 - \theta_3)$. At this point we'll restrict all $ 0 \leq \theta < 2\pi$. So then $\theta_3 - \theta_1 = \theta_1 - \theta_2$. So substituting back into a previous line we get: $e^{i(\theta_2-\theta_1)} + e^{i(\theta_1-\theta_2)} = e^{i\pi}$. Or

$$e^{2i(\theta_2-\theta_1)} + 1 = e^{i(\pi + \theta_2 -\theta_1)} = -e^{i(\theta_2 -\theta_1)}$$ Make the substitution $u = e^{i(\theta_2 - \theta_1)}$ to get $u^2 + u + 1 = 0$. Which means that $e^{i(\theta_2 - \theta_1)} = -\frac{1}{2} \pm \frac{i\sqrt{3}}{2}$. So $e^{i(\theta_2 - \theta_1)} = e^{i\frac{2\pi}{3}}$ or $e^{i\frac{4\pi}{3}}$. In the first case $\theta_2 = \frac{2\pi}{3} + \theta_1$ and $\theta_3 = \frac{4\pi}{3} + \theta_1$ and in the second case $\theta_2$ and $\theta_3$ are reversed; but in both cases $\theta_1,\theta_2,\theta_3$ are the vertices of an equilateral triangle.