Condition for Collinearity of Points in Complex Plane/Formulation 1

Theorem
Let $z_1$, $z_2$ and $z_3$ be points in the complex plane.

Then $z_1$, $z_2$ and $z_3$ are collinear :
 * $\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$

where $\lambda \in \R$ is a real number.

If this is the case, then $z_3$ divides the line segment in the ratio $\lambda$.

If $\lambda > 0$ then $z_3$ is between $z_1$ and $z_2$, and if $\lambda < 0$ then $z_3$ is outside the line segment joining $z_1$ to $z_2$.

Proof
By Geometrical Interpretation of Complex Subtraction:


 * $z_1 - z_3$ can be represented as the line segment from $z_3$ to $z_1$


 * $z_3 - z_2$ can be represented as the line segment from $z_2$ to $z_3$.

Thus we have that $z_1$, $z_2$ and $z_3$ are collinear $z_1 - z_3$ is parallel to $z_3 - z_2$, when expressed as line segments.

Let $\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$ for $\lambda \in \C$.

That is:
 * $z_1 - z_3 = \lambda \left({z_3 - z_2}\right)$

By Complex Multiplication as Geometrical Transformation‎:
 * $\arg \left({z_1 - z_3}\right) = \arg \left({z_3 - z_2}\right) \iff \arg \left({z_3 - z_2}\right) \arg \lambda \in \R_{>0}$

and:
 * $\arg \left({z_1 - z_3}\right) = -\arg \left({z_3 - z_2}\right) \iff \arg \left({z_3 - z_2}\right) \arg \lambda \in \R_{<0}$

where $\arg$ denotes the argument of a complex number.

Also by Complex Multiplication as Geometrical Transformation‎, $z_1 - z_3$ is $\lambda$ the length of $z_3 - z_2$.

The result follows.