I need to show that $\operatorname{Im}(\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1) = 0 \iff z_1,z_2,$ and $z_3$ are collinear.
I know that $\operatorname{Im}(\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1) = 0$ implies that $\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1 \in \mathbb{R}$, but I am not sure how to argue in either direction. Please help. Thank you
Best Answer
We know that $z_1,z_2,z_3$ are collinear iff there exists some $t\in\mathbb{R}$ such that $$t(z_1-z_3)=z_1-z_2.$$ The problem becomes trivial if $z_1=z_3$ so we can suppose this is not the case, and then it is legal to write $$ t=\frac{z_1-z_2}{z_1-z_3}. $$ Multiplying both sides of this equation by $|z_1-z_3|^2=(\overline{z_1-z_3})(z_1-z_3)$ we get \begin{align} t|z_1-z_3|^2=(\overline{z_1-z_3})(z_1-z_2)=&|z_1|^2-\bar z_1z_2 -\bar z_3z_1+\bar z_3z_2\\ =&|z_1|^2-\bar z_1z_2 -\bar z_3z_1\underbrace{-\bar z_2z_3+\bar z_2z_3} _{=0}+\bar z_3z_2.\\ \end{align} Expressing this as $$ t|z_1-z_3|^2-|z_1|^2-(\overline{\bar z_3z_2}+\bar z_3z_2)=-(\bar z_1z_2 +\bar z_2z_3+\bar z_3z_1) $$ and noticing that the LHS is a real number leads to the desired conclusion.