Limitation to "manually" constructing simply connected region in $\mathbb{R}^2$? Restricting the cases to $\mathbb{R}^2$. When studying conservative field, I was under the impression that a vector field can be considered conservative as long as it is continuously differentiable on some region $D$ that is simply connected. 
In situations where the region $D$ is not simply connected (such as a donut with a "hole" in the middle) we can then construct such region into a simply connected one by linking to the boundary of the "hole" two lines with opposite orientations such that they are infinity close to each other, and then extend those two lines to the boundary of the region $D$. Thus transforming $D$ into a simply connected region. 
It follows that I am unsure if there is limitation to this method. Would it be true to state that at least in $\mathbb{R}^2$ any non-simply connected region can be transformed into a simply connected one using the aforementioned method? Would this be true if $D=\mathbb{R}^2$ with a whole in the middle?
As an example, for vector field such as $F=(1/(x^2+y^2), 1/(x^2+y^2)$ which is clearly not defined at the point $(0,0)$. Can I then artificially "remove" the point $(0,0)$ from the whole space $\mathbb{R}^2$ and claim this new region to be simply connected? And thus $F$ is conservative on this region?
Thanks!
P.S. I just made up the vector field $F$, please imagine that it satisfies the requirement to be a conservative field at all points except for $(0,0)$. 
Edit: Sorry to the responders as I do not know a word in topology. My main problem is in fact to justify a problem I saw in my problem set. Here goes:
Let D be the whole region that does not include the point (0,0)
Consider the vector field $f(x,y)=(x/(x^2+y^2), y/(x^2+y^2))$ 
Find a potential for $f$.
Question: Does $f$ even have a potential? Is it even a conservative field? D is clearly not simply connected. I'm sorry if it is really trivial what I fail to see.
Further edit: I think I see where my mistake is. Thanks everyone!
 A: Let me treat the specific question at the end of your post. You are given the vector field
$$ F(x,y) = \left( \frac{x}{x^2+y^2},\frac{y}{x^2+y^2} \right) = (F_1(x,y),F_2(x,y)) $$
on the domain $\Omega = \mathbb{R}^2 \setminus \{ (0,0) \}$. For $F$ to be a conservative vector field, it must have a potential function. A necessary condition for $F$ to be a conservative vector field is $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ and this is also sufficient if the domain $\Omega$ is simply connected. In our case,
$$ \frac{\partial F_1}{\partial y} = -\frac{2xy}{(x^2 + y^2)^2} = \frac{\partial F_2}{\partial x} $$
so the condition is satisfied but since $\Omega$ is not simply connected, we can't deduce that $F$ is conservative. However, we can just try and find a potential function $\phi$ for $F$:
$$ \frac{\partial \phi}{\partial x} = \frac{x}{x^2 + y^2} \implies \phi(x,y) = \int \frac{x}{x^2 + y^2} \, dx + g(y) = \frac{1}{2} \ln (x^2 + y^2) + g(y)$$
where $g$ is some function of $y$. Then
$$ \frac{\partial \phi}{\partial y} = \frac{y}{x^2+y^2} + g'(y) = \frac{y}{x^2 + y^2} \implies g'(y) \equiv 0 $$
so
$$ \phi(x,y) = \frac{1}{2} \ln(x^2 + y^2) + C $$
is a potential for $F$ (where $C \in \mathbb{R}$ is arbitrary). Note that this is well-defined on $\Omega$ (but blows up at the origin).
