Equivalence of two smooth curves in the plane having the same image 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~!

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


*

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

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

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