I would like to prove a proposition which states that three points $z_1$, $z_2$ and $z_3$ are collinear $\iff$
\begin{vmatrix}
z_{1} & \bar{z_{1}} & 1 \\
z_{2} & \bar{z_{2}} & 1 \\
z_{3} & \bar{z_{3}} & 1 \\
\end{vmatrix}=$0$
This is my attempt:
$(\implies)$ Using the first row, multiply $-1$ to it and add it to the second row and obtain:
\begin{vmatrix}
z_{1} & \bar{z_{1}} & 1 \\
z_{2}-z_{1} & \bar{z_{2}}-\bar{z_{1}} & 0 \\
z_{3}-z_{1} & \bar{z_{3}}-\bar{z_{1}} & 0 \\
\end{vmatrix} = $0$ (I have no idea how to typeset equals to zero on the same line as the determinant, my apologies, please bear with me.)
Next, applying cofactor expansion along the third column, we obtain:
\begin{vmatrix}
z_{2}-z_{1} & \bar{z_{2}}-\bar{z_{1}} \\
z_{3}-z_{1} & \bar{z_{3}}-\bar{z_{1}} \\
\end{vmatrix} = $0$
$\implies$ $\bigg(\dfrac{z_3-z_1}{z_2-z_1}\bigg)=\overline{\bigg(\dfrac{z_{3}-z_{1}}{z_{2}-z_{1}} \bigg)}$
$\implies$ $\bigg(\dfrac{z_3-z_1}{z_2-z_1}\bigg)$ is real
$\implies$ $z_1, z_2, z_3$ are collinear.
I feel that my proof is correct, but I help help in conniving myself that the last implication is true. That is, so what if $\bigg(\dfrac{z_3-z_1}{z_2-z_1}\bigg)$ is real?
Best Answer
An alternative approach.
Write $z_j=x_j+iy_j$, then $$\left|\matrix{z_1&\overline{z_1}&1\\ z_2&\overline{z_2}&1 \\z_3&\overline{z_3}&1}\right| =\left|\matrix{x_1+iy_1&x_1-iy_1&1\\ x_2+iy_2&x_2-iy_2&1 \\x_3+iy_3&x_3-iy_3&1}\right|= -2i\left|\matrix{x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1}\right|.$$ The first determinant vanishes iff the last one does, and that occurs iff there is a nonzero column vector with $$\pmatrix{x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1}\pmatrix{a\\b\\c}=\pmatrix{0\\0\\0}.$$ That means that the $(x_j,y_j)$ lie on the line $ax+by+c=0$.