Parametrizatrion of a Straight Line Segment in C - Simple Issue

[Math Amateur]

I am reading Mathews and Howell's book: Complex Analysis for Mathematics and Engineering. I am currently reading Section 1.6 - "The Topology of Complex Numbers". I have a problem with an aspect of Example 1.22 on page 38. My problem is a simple one related to algebraic manipulation of a couple of expressions ...

Example 1.22 reads as follows:

https://www.physicsforums.com/attachments/3747
https://www.physicsforums.com/attachments/3748

In the above M&H write:

" ... ... Clearly one parametrization for -C is

\(\displaystyle -C : \gamma (t) \ = \ z_1 + (z_0 - z_1) t, \ \ \ \text{ for } 0 \le t \le 1\).

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Note that \(\displaystyle \gamma (t) \ = \ z ( 1 - t)\), ... ... ... "

My problem is that I cannot show that \(\displaystyle \gamma (t) \ = \ z ( 1 - t)\), which I suspect is a simple manipulation of symbols ...
Basically, we have

\(\displaystyle \gamma (t) \ = \ z_1 + (z_0 - z_1) t \ = \ z_1 + z_0 t - z_1 t \) ... ... ... ... (1)But \(\displaystyle z (1-t) \ = \ [ z_0 + (z_1 - z_0) t ] ( 1 - t ) \)

\(\displaystyle = z_0 + ( z_1 - z_0 ) t - z_0 t - ( z_1 - z_0 ) t^2\) ... ... ... ... (2)Now I cannot see how further manipulation of the terms of (2) is going to give me the expression in (1) ... ... so how do we derive the equation \(\displaystyle \gamma (t) \ = \ z ( 1 - t)\) ... ...

Hoping someone can help with this (apparently) simple issue ...

Peter

 

[Opalg]

Basically, we have

\(\displaystyle \gamma (t) \ = \ z_1 + (z_0 - z_1) t \ = \ z_1 + z_0 t - z_1 t \) ... ... ... ... (1)But \(\displaystyle z (1-t) \ = \ [ z_0 + (z_1 - z_0) t ] ( 1 - t ) \)

\(\displaystyle = z_0 + ( z_1 - z_0 ) t - z_0 t - ( z_1 - z_0 ) t^2\) ... ... ... ... (2)Now I cannot see how further manipulation of the terms of (2) is going to give me the expression in (1) ... ... so how do we derive the equation \(\displaystyle \gamma (t) \ = \ z ( 1 - t)\) ... ...

Hoping someone can help with this (apparently) simple issue ...

Peter

In this calculation, $z(1-t)$ does not mean $z$ multiplied by $1-t$. It means the function $z$ evaluated at $1-t$. So if $z(t) = z_0 + (z_1 - z_0) t$ then $z(1-t) = z_0 + (z_1 - z_0) (1-t)$.

 

[Math Amateur]

In this calculation, $z(1-t)$ does not mean $z$ multiplied by $1-t$. It means the function $z$ evaluated at $1-t$. So if $z(t) = z_0 + (z_1 - z_0) t$ then $z(1-t) = z_0 + (z_1 - z_0) (1-t)$.

Oh, yes of course!

... ... brain fade on my part ... ... quite silly of me ... ...

Thanks for the help Opalg,

Peter