Let $\alpha, \beta \in \mathbb{C}$. The line segment with end points $\alpha$ and $\beta$ is:
\begin{equation} [\alpha,\beta]:=\{(1-t)\alpha+t\beta : 0\leq t\leq 1\} \end{equation}
How do we explain this? My book just says it is obvious. I kind of get that it might work, but how does someone goes about finding this expression? Clearly we have that at $t = 0$ we get $\alpha$ and at $t = 1$ we get $\beta$. However for $0\leq t \leq 1$ all I see is some kind of weighted sum, but I can't visualize a segment between $\alpha$ and $\beta$. Any suggestion? I'd like to find a way to understand this cause I don't want to just memorize the formula, I want to be able to work it out.
Thank you
We take the straight line passing through $\alpha$ and $\beta$, and parametrise it by
$$\gamma \colon t\mapsto \alpha + t\cdot (\beta - \alpha),\qquad t\in \mathbb{R},$$
choosing $\alpha$ as the base point of the parametrisation and taking the difference between the two points as the direction.
Then, since $\gamma(0) = \alpha$ and $\gamma(1) = \beta$, the line segment between $\alpha$ and $\beta$ is parametrised by the restriction of $\gamma$ to $[0,1]$. We can rearrange the formula for $\gamma(t)$ to get the given form $(1-t)\alpha + t\beta$.