2
$\begingroup$

The book I read is complex analysis by stein, he first define what is the equivalence of two parametrized curve, and then define the complex intagral on a smooth curve which is indepentent of our choice of parametrization. And, my question is about a basic fundamental definition as the following:

If two smooth complex curve $z(t):[a,b]\to C$ and $z_1(s):[c,d]\to C$ have the same image, then does there exists real function $f:[a,b]\to[c,d]$ such that $z(t)=z_1(f(t))$ for all $t$ on $[a,b]$, and $f$ is a continuously differentiable functoin on $[a,b]$, and $f'(x)>0$?

The way of the author use is that if there exist a continuouly differentiable real function $f:[a,b]\to[c,d]$, $f'(x)>0$, and $z(t)=z_1(f(t))=z_1(s)$, then we call the two curve $z(t)$ and $z_1(s)$ are equivalent, but I want to try to claim that it only needs that if two curve have the same image, i.e., $z(t)=z_1(f(t))=z_1(s)$ , and then we can imply the other two requirement to be equivalent.

My attempt: Since $z(t)=z_1(s)=z_1(f(t))$ and $z(t)$ is smooth on $[a,b]$, so $z_1(f(t))$ is continuously differentiable. Next, I guess that we can write $z_1^{-1}$ since $z_1$ is a one to one map, but I don't know how to say that $z_1^{-1}$ is differentiable, if $z_1^{-1}$ is continuously differentiable, then $z_1^{-1}(z_1(f(t)))=f(t)$ is continuously differentiable.

Thanks for any comment and help~!

$\endgroup$
3
$\begingroup$

Hint: Consider the following cases; in each case the images of the curves are the same.

  1. On $ [0,2\pi]:$ $z(t) = e^{it}, z_1(t)= e^{-it}.$

  2. On $ [0,1]:$ $z(t) = t, z_1(t) = t^2.$

  3. On $ [0,2\pi]:$ $z(t) = e^{it}, z_1(t)= e^{2it}.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.