TITLE: Show that three complex numbers $z_1, z_2, z_3$ are collinear iff $\operatorname{Im}(\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1) = 0$
QUESTION [8 upvotes]: 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

REPLY [2 votes]: Here is another approach that relies on the idea that its easy to detect if three complex numbers lie on a line through the origin, and we attempt to reduce the original problem to this simpler one. 
To this end, we might hope that translating our complex numbers (in this case, subtracting $z_1$ from each) doesn't affect whether the function $f(z_1, z_2, z_3) = \overline{z_1}z_2 + \overline{z_2}{z_3} + \overline{z_3}{z_1}$ takes a real value or not. 
And indeed, we'll see that
$$\operatorname{Im}\Big(f(z_1, z_2, z_3)\Big) = \operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big).$$
Observe that
\begin{align*}
\operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big)
&= \operatorname{Im}\Big(\overline{z_2 - z_1}(z_3 - z_1)\Big) \\[7pt]
&= \operatorname{Im}\Big(\overline{z_2}z_3 - \overline{z_1}z_3 - \overline{z_2}z_1 + \overline{z_1}{z_1}\Big) \\[7pt]
&= \operatorname{Im}\big(\overline{z_2}z_3) - \operatorname{Im}\big(\overline{z_1}z_3\big) - \operatorname{Im}\big(\overline{z_2}z_1\big) + \underbrace{\operatorname{Im}\big(|z_1|^2\big)}_{=0}
\end{align*}
where $\overline{z_1}{z_3} = \overline{z_1\overline{z_3}}$ so the two have opposite imaginary parts; that is, $\operatorname{Im}\big(z_1\overline{z_3}) = - \operatorname{Im}(\overline{z_1}z_3)$. Now picking up where we left off,
\begin{align*}
\operatorname{Im}\Big(f(0, z_2 - z_1, z_3 - z_1)\Big)
&= \operatorname{Im}\big(\overline{z_2}z_3) - \operatorname{Im}\big(\overline{z_1}z_3\big) - \operatorname{Im}\big(\overline{z_2}z_3\big) \\[7pt]
&= \operatorname{Im}\big(\overline{z_2}z_3) + \operatorname{Im}\big(\overline{z_3}z_1\big) + \operatorname{Im}\big(\overline{z_1}z_2\big)\\
&= \operatorname{Im}\Big(f(z_1, z_2, z_3)\Big)
\end{align*}
Now we simply need to show that $z_1, z_2, z_3$ are collinear if and only if $f(0, z_2 - z_1, z_3 - z_1) \in \Bbb R$.
Letting $w_2 = z_2 - z_1$ and $w_3 = z_3 - z_1$, this is equivalent to showing that $0, w_1, w_2$ are collinear if and only if $f(0, w_2, w_3) \in \Bbb R$. 
But since $f(0, w_2, w_3) = \overline{w_2}w_3$, this is just the well-known fact that $\overline{w_2}w_3 \in \Bbb R$ if and only if  $w_3 = c w_2$ for some real scalar $c$.