Why is integral not equal to zero even though the path is closed?

Question

I have an integral $$\int{z^{i}}dz$$ the path is $e^{it}$ where $t$ is between $0$ and $2\pi$. Why is it not equal to zero even though the path is closed.

Answer 1

You need to be careful about the definition of $z^i$ as it is singular at $0$. If you insert a branch cut somewhere then the contour is no longer closed. If you do not, you cannot apply Cauchy's theorem.

Answer 2

You seem to believe that $z^i$ has a well-defined meaning, at least as a function from $\mathbb C$ into itself. It hasn't. The number $z^i$ is any number of the form $e^{i\log z}$, where $\log z$ is any logarihtm of $z$ (and, in particular, $z^i$ has no meaning if $z=0$). And you will find no way of choosing, for each $z\in\mathbb{C}\setminus\{0\}$, a number $z^i$ in such a way that the function that you get is continuous, and much less analytic.