Parametrization Of Complex Curves .... Mathews And Howell, Example 1.22 .... ....

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

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.

 

[math771]

Hi Peter,

In this example, we are looking at a complex function z(t) = a + bti, where a and b are real numbers and i is the imaginary unit. The function is defined on the interval 0 ≤ t ≤ 1, which means that t can take on values between 0 and 1.

Now, the function \gamma (t) = z (1-t) is a parametric representation of the same curve as z(t), but in the opposite direction. This means that for each value of t, \gamma (t) will give us a point on the curve in the opposite direction from z(t).

To understand why this is the case, let's look at an example. Let's say we have a point z(0.5) = 2+3i. This means that at t=0.5, our function z(t) will give us the point (2,3) on the complex plane. Now, if we plug in t=0.5 into \gamma(t) = z(1-t), we get \gamma(0.5) = z(1-0.5) = z(0.5) = 2+3i. So, \gamma(0.5) gives us the same point as z(0.5), but in the opposite direction.

In general, we can see that for any value of t, \gamma(t) will give us a point on the curve that is the same distance from the origin as z(t), but in the opposite direction. This is why we can say that \gamma(t) = z(1-t).

I hope this helps clarify the remark made by M&H. Let me know if you have any other questions.