Let $t_1$, $t_2$, $t_3$ be three points on a circle $|t|=1$. If $x_1,x_2,x_2$ are the arguments of the numbers, find $\cos (x_1-x_2)+\cos (x_2-x_3)+\cos (x_3-x_1)$
I have the solution for this question, but I am having trouble understand it. It’s written as
$$|t_1|=|t_2|=|t_3|=1$$
So $$|t_1+t_2+t_3|^2\ge0$$
$$|t_1|^2+|t_2|^2+|t_3|^2+2Re(z_1\bar z_1+z_2\bar z_2+z_3\bar z_3)\ge 0$$
$$3+2(\cos (x_1-x_2)+\cos (x_2-x_3)+\cos (x_3-x_1))\ge 0$$
$$\cos (x_1-x_2)+\cos (x_2-x_3)+\cos (x_3-x_1)\ge \frac{-3}{2}$$
My question is, what is the explanation behind the first expansion of $|t_1^2+t_2^2+t_3^3|$?
Note: I think $z$ is supposed to be $t$. It might be a mistake, but I wasn’t sure so I typed it out as it was.
For any complex number $z, |z|^2 = z\bar z$. So $$\begin{align}|t_1+t_2+t_3|^2 &= (t_1+t_2+t_3)(\bar t_1+\bar t_2+\bar t_3)\\ &= t_1\bar t_1 + t_2\bar t_2+t_3\bar t_3 + (t_1\bar t_2 + t_2\bar t_1) + (t_2\bar t_3 + t_3\bar t_2) + (t_3\bar t_1 + t_1\bar t_3)\\ &=|t_1|^2 + |t_2|^2 +|t_3|^2 + 2\operatorname{Re}(t_1\bar t_2)+ 2\operatorname{Re}(t_2\bar t_3)+ 2\operatorname{Re}(t_3\bar t_1)\\ &= 1 + 1 + 1 +2\left(\cos(x_1 - x_2) + \cos(x_2 - x_3) + \cos(x_3 - x_1)\right) \end{align}$$
Note that $t_1 + t_2 + t_3 = 0$ when the points are equally spaced. I.e. have angles of $\frac {2\pi}3$ between them, so this minimum is obtainable.