Mystery of non-vanishing derivative Studying complex line integrals..
I can't see why we include "non-vanishing derivative" in the definition of a smooth curve.
And although not related -- is complex line integral a type of Riemann-Stieltjes integral?
It seems so but I may miss one point , not so sure..
Thanks in advance..
For example:
Let $z(t)=t^2+it^3$ with $ t\in [0,1]$ then do you say the following is wrong
$$\int_ \Gamma f(z)dz $$ $=$ $$\int_0^1 f(z(t))z'(t)dt $$. 
If not where is the problem exactly? 
Or just visually there will be a cusp but every formula - theorem will still work!?
 A: 
why we include  "non-vanishing derivative" in the definition of a smooth curve.

Not for the sake of having integrals. Integrals work  without this assumption. Having locally Lipschitz parametrization $z(t)$ is enough to make sense of integrals and compute them in the usual way. 
The author of textbook probably assumes this to make some proofs easier: e.g., a proof that the integral $\int_\gamma f $ is independent of how we parametrize $\gamma$ (provided that parametrization runs along it once and in the given direction). 

the definition of a smooth curve.

Also, this definition is not just for integrals. We may want to talk about smooth curves for their own sake, appreciating their elegant shape and subtle curvature. And when doing that, we do not want    to call this curve "smooth":

It has parametrization $z(t) = t^3 + i t^2$. Derivatives of all orders exist... but because of $z'(0)=0$ we get what you see on the plot.


is complex line integral a type of Riemann-Stieltjes integral

It can be defined this way. It can be defined in other ways too. The choice of definition is up to the textbook author. 
