Observe that: $\log(\frac{1+z}{1-z}) = -2\int\frac{dz}{1-z^2}$. (Not precisely true but read on)
Supposedly this function is analytic on the domain $\mathbb{C}-[-1,1]$, despite the fact that it's not decomposable into two analytic functions in the usual way, and thus shouldn't be thought of as a composition of functions.
Now for the closed curve $\alpha (t)=2e^{it}$ for $0\leq t\leq 2\pi$, $\int_{\alpha}\frac{-2dz}{1-z^2} = \int_{\alpha}\frac{-2dz}{(1-z)(1+z)} = 2\pi i$ by the Residue Theorem. Yet by the Cauchy Integral Theorem $\int\frac{-2dz}{1-z^2}$ having a primitive on $\mathbb{C}-[-1,1]$ implies that $\int_{\alpha}\frac{-2dz}{1-z^2} = 0$ for any closed curve on $\mathbb{C}-[-1,1]$.