This is problem 4.1.4 from "Fundamentals of Complex Analysis: with Applications to Engineering and Science".
Problem:
Show that the range of the function $z(t)=t^3+it^6$, $-1\leq t\leq 1$, is a smooth curve even though the given parametrization is not admissible.
There is clearly something I don't understand, since I think that the curve is not smooth. You can find the definition of a smooth curve at the bottom of this post. The condition I can't get my head around is the second one. It says, in my case, $z'(t)$ never vanishes on $[-1,1]$. $z'(t)$ is given by the vector
$$z'(t)=(dx/dt,dy/dt)=(3t^2,6t^5)=3t^2+6t^5i.$$
That it vanishes means that $z'(t)=0$. Hence $$3t^2+6t^5i=0\iff 3t^2=0\land 6t^5=0.$$
The above is true when $t=0$, which is in the specified interval above. So it is not smooth.
I know my conclusion is false since I should prove that the curve is smooth. But I don't get why my reasoning is wrong. :(
Definition:
A point set $\gamma$ in the complex plane is said to be a smooth arc if it is the range of some continuous complex-valued function $z=z(t)$, $a\leq t\leq b$, that satisfies the following conditions:
($i$) $z(t)$ has a continuous derivative on $[a,b]$,
($ii$) $z'(t)$ never vanishes on $[a,b]$,
($iii$) z(t) is one-to-one on $[a,b]$.
A point set $\gamma$ is called a smooth closed curve if it is the range of some continuous function $z=z(t)$, $a\leq t\leq b$, satisfying conditions $(i)$ and $(ii)$ and the following:
$(iii)'$ $z(t)$ is one-to-one on the half-open interval $[a,b)$, but $z(b)=z(a)$ and $z'(b)=z'(a)$.
That "$\gamma$" is a smooth curve" means that $\gamma$ is either a smooth arc or a smooth closed curve.
Thanks for your time! :)