Let $\Omega \subseteq \mathbb{C}$ be open and $\gamma:[\alpha,\beta]\rightarrow \Omega$ be a piecewise continuously differentiable and closed path.
Why does $z^{-1}$ have no antiderivative on $\mathbb{C}\setminus0$?
Why is $\int\limits_\gamma z^{-1} dz = 2\pi i \cdot \text{ind}_\gamma (0)$, where $\text{ind}_\gamma$ denotes the winding number.
Best Answer
Potato's answer gives you all you need. But I'd like to mention not why these facts are true, but why they should be true. So here's how I like to think about it (all of which can be made precise):
We know that the derivative of the natural log function is $z^{-1}$, so locally the antiderivative of $z^{-1}$ is $\ln$. But since the exponential function is $2\pi i$-periodic, $\ln$ is not well-definied: if we start by defining, say, $\ln(1)$, we have to choose which inverse function to use: should that be $0$, or $2\pi i$, or $-38\pi i$? So we make a choice, and continuing along the unit circle, we define natural log to be the function whose anti-derivative is $z^{-1}$ which is consistent with our original choice. (This can be made precise, and is an example of what is called analytic continuation.) Since the imaginary part of $\ln$ measures argument (angle), an we have been moving in the positive direction the whole time, we see that our function is not consistent: when we get back to $1$ we are $2\pi i$ more than where we started! So although we can locally define $\ln$, we run into trouble trying to define the function on all of $\mathbb{C}\backslash \{0\}$. This same inconsistency, via the fundamental theorem of calculus, explains the residue formula as well: one bound of the integral is given by one choice of $\ln(1)$, and the other bound is evaluated using another choice.