MHB 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

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

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

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

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

$$= z_0 + ( z_1 - z_0 ) t - z_0 t - ( z_1 - z_0 ) t^2$$ ... ... ... ... (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 $$\gamma (t) \ = \ z ( 1 - t)$$ ... ...

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

Peter

 

[Opalg]

Basically, we have

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

$$= z_0 + ( z_1 - z_0 ) t - z_0 t - ( z_1 - z_0 ) t^2$$ ... ... ... ... (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 $$\gamma (t) \ = \ z ( 1 - t)$$ ... ...

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