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
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.
The basic concept for this answer is the area of a triangle. The area of a triangle which is made on vectors $z$ and $w$ is $A=\dfrac12{\bf Im}(\bar{z}w)$. for three points in complex plane, we make a triangle on them with sides $z_1-z_2$, $z_2-z_3$ and $z_3-z_1$, we have \begin{eqnarray} \dfrac12{\bf Im}\overline{(z_1-z_2)}(z_2-z_3)&=&0\\ {\bf Im}\overline{z_1}z_2+\overline{z_2}z_3-\overline{z_2}z_2-\overline{z_1}z_3&=&0\\ {\bf Im}\overline{z_1}z_2+\overline{z_2}z_3+{\bf Im}\overline{z_3}z_1&=&{\bf Im}\overline{z_2}z_2+\overline{z_1}z_3+{\bf Im}\overline{z_3}z_1\\ {\bf Im}\overline{z_1}z_2+\overline{z_2}z_3+\overline{z_3}z_1&=&0 \end{eqnarray}
$z_1$, $z_2$ and $z_3$ are collinear, if and only if, the area of triangle made on them is zero.
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 $c$.