I've been trying to prove that this vector field:
$$
\vec{F}=\left(\frac{y}{\left(x-1\right)^{2}+y^{2}},\frac{1-x}{\left(x-1\right)^{2}+y^{2}}\right)
$$
Is conservative in:
$$
D=\left\{ \left(x,y\right)\mid1\leq\left(x+1\right)^{2}+y^{2}\leq2\right\}
$$
Proof attempt #1:
To my understanding, it holds that in a simply-connected domain, a vector field $P\hat{x}+Q\hat{y}$ whose partial derivatives are continuous is conservative if and only if $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$.
I confirmed that in the circle centered at $\left(-1,0\right)$ with a radius of $2$, which is a superset of $D$, the partial derivatives are indeed continuous, and furthermore that:
$$
\frac{\partial Q}{\partial x} = \frac{\left(x-1\right)^{2}-y^{2}}{\left(\left(x-1\right)^{2}+y^{2}\right)^{2}} = \frac{\partial P}{\partial y}
$$
So in that circle, $\vec{F}$ is indeed conservative.
Thus: for any closed, piecewise-smooth curve in that circle, we have that the line integral of $\vec{F}$ over that curve is $0$. Since any such curve in $D$ is also a curve in the circle (and since adding that hole-of-radius-1 in the middle of it doesn't affect any of the components of a line integral: the field in all points on the curve and any parametrization of the curve, then it also doesn't affect the line integral itself), we have that every such curve in D also has that property, and so $\vec{F}$ is conservative there, as well.
Does the above proof seem okay (if I make it a tad more rigorous, maybe)?
Proof attempt #2:
Another possible proof method that I'd considered might be using Green's Theorem to show that the line integral over the unit circle centered at $\left(-1,0\right)$ results in $0$, and then use a theorem (which I believe I can prove) by which if a field has the aforementioned property of $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$ (and probably a few other requirements) then any closed, simple curve around such a "hole" in its domain would result in the exact same value, so not exactly conservative because that value doesn't have to be $0$, but since I already showed that one such integral does evaluate to $0$, then all of them do, so the field is indeed conservative.
Does this one make sense?