I have been asked to show that the function $f(z) = \frac{z}{z^2+1}$ does not have an antiderivative on the subset $\{z \in \mathbb{C}: 1 < |z| \}$. However I don't see why this is the case, we can write $\frac{z}{z^2+1} = \frac{1}{2}(\frac{1}{z+i}+\frac{1}{z-i})$ whose naive antiderivative (using $\ln$) has branch points at $\pm i$ so taking a straight line between these points as our branch cut should allow us to make it continuous and single valued on our region. What am I missing?
I know I am making a mistake somewhere as the residue theorem tells me the integral around a circle of radius 2 has value $2\pi i$ and not $0$ so no antiderivative exists. Any help is much appreciated.
Consider a positively oriented circle $\Gamma$ of radius $ > 1$ centred at the origin.
$$ \dfrac{1}{2\pi i} \oint_\Gamma \dfrac{z}{z^2+1} dz$$ is the sum of the residues of $z/(z^2+1)$ at $\pm i$, namely $1$. But if your function had an antiderivative on that region, the result would have to be $0$.