Condition for Collinearity of Points in Complex Plane/Formulation 2

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Theorem

Let $z_1, z_2, z_3$ be distinct complex numbers.


Then:

$z_1, z_2, z_3$ are collinear in the complex plane

if and only if:

$\exists \alpha, \beta, \gamma \in \R: \alpha z_1 + \beta z_2 + \gamma z_3 = 0$
where:
$\alpha + \beta + \gamma = 0$
not all of $\alpha, \beta, \gamma$ are zero.


Proof

Sufficient Condition

Let $z_1, z_2, z_3$ be collinear.


Then by Condition for Collinearity of Points in Complex Plane: Formulation 1 there exists a real number $b$ such that:

$z_2 - z_1 = b \paren {z_3 - z_1}$

Then:

\(\ds z_2 - z_1\) \(=\) \(\ds b \paren {z_3 - z_1}\)
\(\ds \leadsto \ \ \) \(\ds z_2 - z_1 - b z_3 + b z_1\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {b - 1} z_1 + z_2 - b z_3\) \(=\) \(\ds 0\)

Setting $\alpha = b - 1, \beta = 1, \gamma = -b$ fits the bill, as $\paren {b - 1} + 1 + \paren {-b} = 0$.

$\Box$


Necessary Condition

Let $\alpha + \beta + \gamma = 0$ such that:

$\alpha z_1 + \beta z_2 + \gamma z_3 = 0$
at least one of $\alpha, \beta, \gamma$ is not zero.


Without loss of generality let $\alpha \ne 0$.

Then it follows that as $\alpha + \beta + \gamma = 0$, at least one of $\beta$ and $\gamma$ is also non-zero.

Without loss of generality let $\beta \ne 0$.

In the following it is immaterial whether $\gamma = 0$ or not.

We have:

\(\ds \alpha + \beta + \gamma\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \gamma\) \(=\) \(\ds -\paren {\alpha + \beta}\)
\(\ds \leadsto \ \ \) \(\ds \alpha z_1 + \beta z_2\) \(=\) \(\ds \paren {\alpha + \beta} z_3\) from $\alpha z_1 + \beta z_2 + \gamma z_3 = 0$
\(\ds \leadsto \ \ \) \(\ds z_1 + \frac \beta \alpha z_2\) \(=\) \(\ds \frac {\alpha + \beta} \alpha z_3\) which can be done because $\alpha \ne 0$
\(\ds \leadsto \ \ \) \(\ds z_1 + \frac \beta \alpha z_2 - \paren {1 + \frac \beta \alpha} z_1\) \(=\) \(\ds \frac {\alpha + \beta} \alpha z_3 - \paren {1 + \frac \beta \alpha} z_1\)
\(\ds \leadsto \ \ \) \(\ds \frac \beta \alpha \paren {z_2 - z_1}\) \(=\) \(\ds \frac {\alpha + \beta} \alpha \paren {z_3 - z_1}\) simplifying
\(\ds \leadsto \ \ \) \(\ds z_2 - z_1\) \(=\) \(\ds \frac \alpha \beta \frac {\alpha + \beta} \alpha \paren {z_3 - z_1}\) multiplying both sides by $\dfrac \alpha \beta$, which can be done because $\beta \ne 0$
\(\ds \leadsto \ \ \) \(\ds z_2 - z_1\) \(=\) \(\ds \frac {\alpha + \beta} \beta \paren {z_3 - z_1}\)

Thus it is seen that:

$z_2 - z_1 = b \paren {z_3 - z_1}$

for some $b \in \R$.

Hence by Condition for Collinearity of Points in Complex Plane: Formulation 1, $z_1$, $z_2$ and $z_3$ are collinear.

$\blacksquare$


Sources

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