What can we say about $ b\overline {z}+\overline {b}z$

Question

What can we say about $\overline {b}\left( z^{2}_{1}+z^{2}_{2}+z^{2}\right) +b\left( \overline {z_{1}}^{2}+\overline {z_{2}}^{2}+\overline {z_{3}}^{2}\right) $, if $b, z_{i}\in \mathbb{C} ,\left| z_{i}\right| =1$ and $A\left( z_{1}\right) ,B\left( z_{2}\right) ,C\left( z_{3}\right) $ are the afixes of an equilateral triangle?

Answer 1

Let $w_1=z_1z_3^{-1}$ and $w_2=z_2z_3^{-1}$. Then, as we are simply rotating the triangle by the angle determined by $z_3$, we have $w_1=\omega$ and $w_2=\omega^2=\bar{\omega}$, the nonreal cube roots of $1$. Thus $\omega^2+\omega+1=0$ and $\omega^4=\omega$.

Your expression then becomes $$ \bar{b}z_3^2(\omega^2+\omega^4+1)+b\bar{z}_3^2(\bar{\omega}^2+\bar{\omega}^4+1) $$

Answer 2

Since $$\overline{w}=w\Leftrightarrow w\in\mathbb R,$$ we can say that it's a real number.