Need help to understand the proof of collinearity of 3 complex numbers

Question

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?

Answer 1

Consider the function $$ f(z) = \begin{vmatrix} z_{1} & \bar{z_{1}} & 1 \\ z_{2} & \bar{z_{2}} & 1 \\ z & \bar{z} & 1 \\ \end{vmatrix}. $$ First of all, we have $\overline{f(z)} = -f(z)$, since taking the complex conjugate of the entries will simply swap the first and second column. So $f(z)$ is purely imaginary.

So setting $f(x+yi) = 0$ is a single linear equation in $x$ and $y$. (Linear in the affine sense: it has a constant term.) It's linear in $x$ and linear in $y$ because we never multiply those terms when taking the determinant. It's only one equation (not two) because the real part is guaranteed to be $0$, no matter what; we are only checking that the imaginary part is $0$.

(Exceptional case: $f(x+yi)=0$ for all $x,y$, imposing no conditions at all, exactly when $z_1 = z_2$.)

Therefore the graph of $f(z) = 0$ is a line in the complex plane. What's more, we have $f(z_1) = f(z_2) = 0$, because setting $z = z_1$ or $z = z_2$ gives us a matrix with two equal rows, whose determinant is $0$. So it's a line passing through $z_1$ and $z_2$.

As a result, $f(z_3) = 0$ if and only if $z_3$ is on the line through $z_1$ and $z_2$, which is just what we wanted to know.

Answer 2

$$\bigg(\dfrac{z_3-z_1}{z_2-z_1}\bigg)=\lambda \implies z_3-z_1=\lambda (z_2-z_1)$$ Which means the vectors $z_1 z_3$ has the same direction as $z_1z_2$ hence collinearity of three points is verified.

Answer 3

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 $ax+by+c=0$.