Why does this not contradict Cauchy's theorem?

Question

Let $D=\{z \in \mathbb{C}\ : 1<|z|<3\}$, and put $f(z)=1/z$. Let $\alpha:[0,1] \rightarrow 2\exp(2i\pi t)$. Then $\int_\alpha f \, dz=2\pi i$. But surely Cauchy's theorem says that this integral is $0,$ since $f$ is analytic in $D$?

Answer 1

It turns out that $\alpha$ is not null-homotopic in $D$.

Answer 2

Wikipedia's account of Cauchy's theorem says:

[L]et $U$ be an open subset of $\mathbb C$ which is simply connected, let $f:U\to\mathbb C$ be a holomorphic function, and let $\gamma$ be a rectifiable path in $U$ whose start point is equal to its end point. Then $$\oint_\gamma f(z)\,dz = 0. $$

The open set that you describe is not simply connected (although it is connected), so Cauchy's theorem is not applicable.