Prove $z_1$, $z_2$ ,$z_3$ on complex plane are collinear iff $\frac{z_3-z_1}{z_2-z_1}\in\mathbb R$

Question

Prove that complex numbers $z_1$, $z_2$ ,$z_3$ on complex plane are collinear iff $\displaystyle\frac{z_3-z_1}{z_2-z_1}\in\mathbb R$

I think if $\displaystyle\frac{z_3-z_1}{z_2-z_1}\in\mathbb R$ then $\displaystyle\frac{z_3-z_1}{z_2-z_1}$ lies on the real axis.

So, $\displaystyle\arg\left(\frac{z_3-z_1}{z_2-z_1}\right)=\arg(z_3-z_1)-\arg(z_2-z_1)=0$ or $\pm\pi$.

Can I conclude $z_1, z_2, z_3$ are collinear? and if $z_1, z_2 ,z_3$ on complex plane are collinear. How to prove $\displaystyle\frac{z_3-z_1}{z_2-z_1}\in\mathbb R$ Please help!

Answer 1

Think of complex numbers as position vectors. For example, $2 + 3i$ can be thought of as the position vector of the point $(2,3)$ in the coordinate plane. Following this line of reasoning, you can think of $z_1 - z_3$ as a vector starting at $z_3$ and pointing to $z_1$. In other words, $z_1 - z_3$ is the displacement vector if you travel from $z_3$ to $z_1$.

Now if $z_1, z_2, z_3$ are collinear, then the vectors $z_1 - z_3$ and $z_1 - z_2$ must be parallel (or antiparallel) and thus $z_1 - z_3$ = $ \lambda (z_1 - z_2)$ for some $\lambda \in \mathbb{R}$.

Similarly, if $\frac{z_1 - z_3}{z_1 - z_2} = \lambda$ then the vectors $z_1 - z_3$ and $z_1 - z_2$ must be parallel (or antiparallel) and therefore the points $z_1, z_2, z_3$ must be collinear.

As you can see, thinking of complex numbers as vectors can sometimes provide excellent intuition into a seemingly hard problem.

Answer 2

Let $z_3-z_1=a e^{i\alpha}$ and $z_2-z_1=b e^{i\beta}$ to express

$$\frac {z_3-z_1}{z_2-z_1} = \frac {ae^{i\alpha}}{be^{i\beta}} =\frac abe^{i(\alpha-\beta)} =\frac ab [ \cos(\alpha-\beta)+i\sin(\alpha-\beta)] $$

Thus,

$$\frac{z_3-z_1}{z_2-z_1}\in\mathbb R\iff \sin(\alpha-\beta)=0\iff\arg(z_3-z_1)-\arg(z_2-z_1)=n\pi$$