Show that $ABC$ is equilateral if and only if $a+b+c=0$

Question

In the complex plane let $a,b,c \in \Bbb C^3$ represent the points $A,B,C$ all three different from the origin $O$ . Show that $ABC$ is equilateral if and only if $$ a+b+c=0$$ I found this in an exam paper and I'm not even sure if it's correct and I really don't know how to move forward after establishing the obvious characteristics of the equilateral triangle. Any help is greatly appreciated!

Answer 1

Note that we need that $|a|=|b|=|c|$ as extra condition otherwise we can find solutions which do not represent equilateral triangle (e.g. $a=1, b=c=-\frac12$).

Assuming $|a|=|b|=|c|$, we have that for an equilateral triangle

• $a=re^{i\theta}$
• $b=re^{i\theta+i\frac {2\pi} 3}$
• $c=re^{i\theta+i\frac {4\pi} 3}$

therefore

$$a+b+c= re^{i\theta}(1+e^{i\frac {2\pi} 3}+e^{i\frac {4\pi} 3})=re^{i\theta}\cdot 0=0$$

For the other direction, assuming wlog, up to scaling and rotation, $a=1 \implies b+c=-1$ that is

• $a=1$
• $b=-\frac12+iy$
• $c=-\frac12-iy$

and since $|b|=|c|=1$ we obtain

$$\sqrt{\frac14+y^2}=1 \implies y^2=\frac34 \implies y=\frac{\sqrt 3}2$$

Answer 2

As I said in the comment the statement is false in general. But a if you assume that $a, b, c$ have the same norm it become true. In fact WLOG you can assume $a=1$, $b=e^{i\beta} $ and $c=e^{i\gamma} $, this because you can first divide by the norm of $a,b$ and $c$ and then rotating the plan such that $a\to 1$. Now we have that $1+e^{i\beta}+e^{i\gamma}=0$ that is $\begin{cases} 1+\cos\beta +\cos \gamma=0\\ \sin\beta+\sin\gamma =0\end{cases}$. The second condition means $\gamma +\beta =\pi$ and I'm sure you are able to conclude!