Let the points \(a\), \(b\), and \(c\) lie on the unit circle of the complex plane and satisfy \(a+b+c=0\). Prove that \(a\), \(b\), and \(c\) form the vertices of an equilateral triangle.
Solution
1. One has\[ \operatorname{Re}(a\overline b) = \operatorname{Re}(b\overline c) = \operatorname{Re}(c\overline a) = -\frac12. \]
Since \(a+b=-c\) and \(\abs{a}=\abs{b}=\abs{c}=1\), \[ 1 = \abs{a+b}^2 = (a+b)(\overline a+\overline b) = 2+2\operatorname{Re}(a\overline b). \] Hence \[ \operatorname{Re}(a\overline b)=-\frac12. \] The same argument applied to \(b+c=-a\) and \(c+a=-b\) gives the other two identities.
2. The three pairwise distances are equal:\[ \abs{a-b} = \abs{b-c} = \abs{c-a} = \sqrt3. \]
Using step 1, \[ \begin{aligned} \abs{a-b}^2 &= (a-b)(\overline a-\overline b)\\ &= 2-2\operatorname{Re}(a\overline b)\\ &= 3. \end{aligned} \] The identical calculation for the pairs \((b,c)\) and \((c,a)\) gives squared distance \(3\) in each case.
3. The points \(a,b,c\) are the vertices of an equilateral triangle.
By step 2 all three side lengths are equal to \(\sqrt3>0\). In particular the points are distinct, and the triangle they determine is equilateral.
4. Q.E.D.
Step 3 is the required conclusion.