Show that a complex number is on a line.

Question

Let $v$ and $w$ be two distinct complex numbers such that $v + t (w-v)$ is a line $l$, where $t \in \mathbb{R}$. Prove that:

If $\frac{z - w}{z-v}$ is a real number, for instance $t$, then $z$ is on the line $l$.

I have already tried using the hint as follows:

\begin{equation} \frac{z - w}{z-v}=t \end{equation} \begin{equation} z - w=t(z-v) \end{equation} \begin{equation} z = \frac{-v t +w}{1-t} \end{equation}

This doesn't really get me anywhere, at least I think it doesn't because I don't recognise the form of my desired line in here. I also tried using the following algorithm: \begin{equation} \frac{z - w}{z-v}(w-v)=t(w-v) \end{equation} \begin{equation} \frac{z - w}{z-v}(w-v)+v=v+ t(w-v) \end{equation} Which can be rewritten as: \begin{equation} \frac{zw-w^2 +vw-v^2}{z-v}=v+ t(w-v) \end{equation}

I would have hoped it to simplify to $z$. Do you people have any pointers or tips?

Answer 1

Your relation $$ z = \frac{-v t +w}{1-t} $$ can be written as $$ z=v+\tau(w-v), $$ where $\displaystyle\tau={1\over1-t}={z-v\over w-v}$.

Answer 2

Hint for alternate solution: Think geometrically. We have that $l$ is the line going through the points $v$ and $w$ in the complex plane. What can you say about two complex numbers (interpreted as vectors in the plane) if their ratio (as complex numbers) is real?

Answer 3

Just one more step.

$$z=\frac{-vt+w}{1-t}=v+\frac{w-v}{1-t}=v+t'(w-v)$$

with $t'=\frac{1}{1-t}$.