Is this vector field conservative? I'm considering a vector field $\mathbf{F}$ defined by a continuosly differentiable function from $A=\{(x,y) \in \mathbb{R}^2 : x^2+y^2>1\}$ into $\mathbb{R}^2$. 
I know that the vector field is associated to a closed differential form, meaning that the field is irrotational.
If I suppose that $\mathbf{F}(x,y) = (0,0)$ for every $(x, y) \in \mathbb{R}^2$ such that $x^2+y^2=2$, I have a closed curve $\gamma$ such that $\int_{\gamma} \mathbf{F} \cdot d\mathbf{l} = 0$. Then, for every closed curve $\gamma_1$ homotopic to $\gamma$ I should have $\int_{\gamma_1} \mathbf{F} \cdot d\mathbf{l} = 0$. Can I conclude that the field is conservative over $A$?
I think the answer is no, because the above only holds for closed curves that are homotopic to $\gamma$, i.e. closed curves containing the "hole" in the region A (circle of radius 1) while curves whose interior doesn't contain the hole could produce different values for the line integral, but I'm not convinced about this. Is it correct? Could someone give me an example field where this holds?
 A: Actually, you can conclude that the vector field is conservative (i.e the line integral is path independent). This is because your region $A$ is an (infinitely large) annulus, so two continuous closed curves $\gamma_1, \gamma_2$, with image lying inside $A$ are (continuously) homotopic as closed curves in $A$ if and only if their index with respect to the origin are equal:
\begin{align}
I(\gamma_1, 0) &= I(\gamma_2, 0)
\end{align}
where the index (also called winding number I think) is defined as (using a little complex analysis)
\begin{align}
I(\gamma_1, 0):= \dfrac{1}{2\pi i}\displaystyle\int_{\gamma_1} \dfrac{dz}{z}
\end{align}
The index is always an integer. Assuming you take these facts for granted, we can proceed as follows. Let $\gamma$ be any continuous closed path in $A$, and let $n\in \Bbb{Z}$ be the index. Note also that the path $\delta_n:[0,2\pi] \to A$ defined by $\delta_n(t) = 2e^{int}$ has index $n$, hence there is a homotopy (within $A$) between $\gamma$ and $\delta_n$. Since $\mathbf{F}$ arises from a closed differential form, we have
\begin{align}
\int_{\gamma}\mathbf{F} \cdot \mathbf{dl} &= \int_{\delta_n}\mathbf{F} \cdot \mathbf{dl} = 0,
\end{align}
where the last equal sign is because the image of $\delta_n$ lies inside the set where $\mathbf{F}$ vanishes. 
Hence, we have shown that the integral of the vector field over every closed path vanishes, which means it is conservative.

By the way, note that your intuitive reasoning is incorrect, because even if the path doesn't contain the hole, it can be continuously deformed to a point (because both these paths have index zero... but I'm sure you can also convince yourself pictorially). So, in that case the integral will still be zero.
As for examples, of course the zero vector field works. But more generally, for a non-trivial example, take the vector field
\begin{align}
\mathbf{F}(x,y) &= \bigg(4x(x^2 + y^2 -2), 4y(x^2 + y^2 - 2)\bigg)
\end{align}
This clearly vanishes on the circle $x^2 + y^2 =2$. It is also a conservative vector field, because it is the gradient of $f(x,y) = (x^2 + y^2 - 2)^2$ (in coming up with this example, I of course started with $f$ and then calculated $\mathbf{F} =\nabla f$)
