Suppose $f$ is analytic on $\Omega$, consider $\int_{\tau} f\,dz$, where $\tau$ is given by $\tau(t) = \cos(t) + i\sin(t)$ for $0 \leq t \leq \pi$, and by $\tau(t) = \cos(t) - i\sin(t)$ for $\pi \leq t \leq 2\pi$.
If we draw the graph we can see that between the the graph will be a arc in the counter clock wise direction for $t \in [0,\pi]$, i.e it will be a half circle, while for $t \in [\pi,2\pi]$ the circle will start at $2\pi$ and end at $\pi$, so is the integral zero by Cauchy integral theorem ?
since $\tau$ on $[\pi, 2\pi]$ simply reverses its course on $[0,\pi]$, it doesn't even require the Cauchy integral theorem, or even for $f$ to be analytic. All that is required is that $f$ be integrable along $\tau$. We have $\int_\tau f\ dz = 0$ because $$\int_0^{\pi} f\circ \tau(t) \tau'(t)dt = -\int_{\pi}^{2\pi} f\circ \tau(t) \tau'(t)dt$$