Let $\gamma: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to \mathbb C$ be given by $\gamma (t) = 2e^{it}$. Calculate $$\int_{\gamma} \frac{z}{z^2 + 1} dz$$ by finding area $U$ containing $\gamma^* = \gamma \left( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \right) $ and an antiderivative $F$ of function $f(z) = \frac{z}{z^2 + 1}$ on area $U$. Use the equality: $\int_{\gamma} f(z) dz = F(\gamma(\frac{\pi}{2})) - F(\gamma(-\frac{\pi}{2})).$
Now I can find the antiderivative. It is just $\frac{\ln(z^2 + 1)}{2}$. The problem is that $\gamma (t) = 2e^{it}$ is a ring around $0$ and there is no way to cover that area with logarithm. So I need to somehow merge more than one area to get my $U$. I don't know how to do that.