MHB Why Does γ(t) = z(1-t) Represent the Same Curve in Reverse?

[Math Amateur]

I am reading "Complex Analysis for Mathematics and Engineering" by John H. Mathews and Russel W. Howell (M&H) [Fifth Edition] ... ...

I am focused on Section 1.6 The Topology of Complex Numbers ...

I need help in fully understanding a remark by M&H ... made just after Example 1.22 ...

Example 1.22 (including some post-Example remarks ...) reads as follows:View attachment 7342
View attachment 7343In the above text from Mathews and Howell we read the following in remarks made after the presentation of the example:

" ... ... Note that $$\gamma (t) = z ( 1-t )$$, ... ... "Can someone please explain how/why $$\gamma (t) = z ( 1-t )$$ ... ... ?

Peter

 

[Euge]

You could just start from $z(t) = z_0 + (z_1 - z_0)t$, replacing $t$ by $1 - t$ and work out the algebra to get the expression $z_1 + (z_0 - z_1)t$. Alternatively, you can view this conceptually. The map $t \mapsto 1 - t$ from $[0,1]$ to $[0,1]$ maps $0$ to $1$ and $1$ to $0$, so $z(1-t)$ traces $C$ in the opposite direction that $z(t)$ does.