# Points on a straight line (Complex Analysis)

I encouter a problem in complex analysis course :

Let $a, b,$ and $c$ be three distinct points on a straight line with $b$ between $a$ and $c$. Show that $\frac{a-b}{c-b} \in \mathbb{R}_{<0}$.

Precisely, I am not sure how to begin. Generally, I think that a point in complex plane $z = z + iy$ can be identified using coordinate $(x,y)$. But this seems does not make sence since I need to show that the ratio of different belongs to the set of negative real number, which is clearly in one dimension. Alternatively, I think that I might let $a = a_1 + ia_2, b = b = b_1 + i b_2$ and $c = c_1 + i c_2$. But I do not sure what form the straight line shold be. Is it $z_2 = az_1 + b$ where $z_1, z_2 \in \mathbb{C}$ and $a, b$ are complex constants? Please hint or give some useful suggestion. Thank you very much.

Let the line be $L=\{z_0+t d_0\}_{t \in \mathbb{R}}$.
Then $a=z_0+t_a d_0$, $b=z_0+t_b d_0$ and $c=z_0+t_c d_0$ for some distinct reals $t_a,t_b,t_c$.
Then $a-b = (t_a-t_b) d_0$, $c-b = (t_c-t_b) d_0$, and so we have ${a-b \over c-b} = { t_a-t_b \over t_c-t_b} \in \mathbb{R}$.
• Here $z_0$and $d_0$ are complex, right? I am not sure about the form of straight line. As you do, I suppose that any straight line in complex take form as your $L$. I think that, according to the form of $L$, it seems that the line equation is parametized by $t$. Is it something like representing lines in vector calculus (using a vector that the line parallel to, and parametrized by $t$). Anyway, thank you very much. – Both Htob Jan 29 '15 at 5:49
• Yes, $z_0$ is a point on the line and $d_0$ is the direction. I don't understand your next sentence. It is much the same as a line in $\mathbb{R}^2$ with the exception that we can divide points in $\mathbb{C}$. – copper.hat Jan 29 '15 at 5:52
If $z$ is on a straight line going through $z_0$ with angle $\theta$ then $z = z_0 + te^{i\theta}$ for some $t \in \mathbb{R}$