This is a lot of work/questions, so I do apologize, I only come here when I am in desperate need of help and I may have to rewatch a lot of lectures.
-Does a gradient field need to be defined on the entire plane? If I have a hole at the origin/the region isn't simple connected, does that mean it can't possibly be a gradient field? And if I have a hole at the origin, the vector field is not conservative?
-The vector field \frac{xi+yj}{x^2 + y^2} has a hole at the origin but the line integral is 0, showing that it is conservative. How is that possible?
-Is it possible for a vector field to be a gradient field in one part of the plane but a non gradient field on another part? Or is something either just a gradient vector field or just a non gradient vector field? The vector field \frac{-yi+xj}{x^2 + y^2}has a line integral around the origin evaluate to 2pi, that proves that the entire field is not a gradient field right?
-If I have a gradient field with a hole in the origin, does that mean it can be conservative on one part of the field, but not on another (like the part that encloses the origin)?
Any help is greatly, greatly appreciated
$1$. This question is "backwards" in logic, because one ALWAYS has to specify a domain of definition before even beginning to ask things like "is this conservative/non-conservative" or "is this a smooth vector field", or anything else.
$2$. The vector field you consider is the gradient of $f: \Bbb{R}^2 \setminus \{0\} \to \Bbb{R}$ defined by $f(x,y) = \dfrac{1}{2}\ln(x^2 + y^2)$. So, the line integral is path-independent.
$3$. If you allow the domain of the vector field to be disconnected, then all sorts of things are possible. Yes, the vector field $\xi: \Bbb{R}^2 \to \Bbb{R}^2$ defined by $\xi(x,y) = \left(\dfrac{-y}{x^2 + y^2}, \dfrac{x}{x^2 + y^2} \right)$ is not conservative, because its integral over a circle around the origin (oriented counter-clockwise) is $2 \pi$.
$4$. (I'm not really sure what you mean for your last paragraph.)