Reasoning behind Fields with "holes" For this vector field
$$F(x, y) = (-\frac{y}{(x+1)^2+y^2} - \frac{y}{(x-1)^2+y^2}, \frac{x+1}{(x+1)^2+y^2} + \frac{x-1}{(x-1)^2+y^2})$$
I'm asked to check if it is a gradient field in region $D = \{(x, y): 4 \leq x^2 + y^2 \leq 9\}$.
So, $D$ is a ring of that goes from radius 2 to radius 3. $F$, in the other hand, is undefined at $(-1, 0)$ and $(1, 0)$. I thought that, because those points didn't belong to $D$, then the field $F$ would be conservative at $D$ where the singularities are not defined.
But the solutions in the textbook say that, because those points are "holes", we can conclude that the work done by $F$ along any circumference (positively defined) centered at the origin is $4\pi$. So it is not a gradient field.
I'm very confused by this answer and not quite grasping it.

*

*The points are not in $D$. So why are they being considered to the work the field does in $D$?

*If I parametrise the inner circle of the ring (counterclockwise) and the outer circle (clockwise), don't both lines cancel each other by a factor of $2\pi$? I would guess that even with that, there is the "remaining" $4\pi$ from the "holes" within?

*If there were no "holes" would the field be conservative in $D$? If there were only one "hole", would F not be conservative by a factor of $2\pi$?

*So every field with a "hole" in it is not conservative - but if there was a "hole counterclockwise" (does that exist?) they would cancel each other out and the field would already be conservative?

 A: tl; dr: Whether or not a plane vector field is gradient depends on the domain of the field, not merely on the defining formula. (Let that sink in for a moment!) "Holes" in the domain are related, but are not (as it were) the whole story.

To address the question definitively, it's worth taking a step back. Let $R$ denote the punctured plane, and consider the vector fields
$$
A(x, y) = \frac{(-y, x)}{x^{2} + y^{2}},\qquad
B(x, y) = \frac{(x, y)}{x^{2} + y^{2}}
$$
in $R$. Each field has (scalar) curl $0$, so each is "locally gradient." Concretely, if $D$ is an open disk contained in $R$, then each field is gradient in $D$. (There is nothing special about disks here, however. Instead, $D$ could be an open rectangle contained in $R$, or a simply-connected open subset of $R$, or merely a subset of a simply-connected open subset of $R$, such as an annulus not enclosing the origin.)
Separately, the field $B$ is gradient in $R$: If $b(x, y) = \frac{1}{2}\ln(x^{2} + y^{2})$, then $B = \nabla b$ throughout $R$. Consequently, $B$ is a gradient field when restricted to an arbitrary non-empty open subset of $R$.
By contrast, the integral of $A$ counterclockwise around the unit circle is $2\pi$, while the integral of a gradient field is $0$ by the fundamental theorem of calculus. We conclude that $A$ is not a gradient field in $R$. To emphasize, $A$ is gradient in every open disk contained in $R$, but is not gradient in all of $R$. To reiterate the key conceptual point of the question: Whether or not a plane vector field is gradient depends on the domain of the field, not merely on the formula defining the field.
In this example we have a pleasant geometric interpretation: Formally, $A$ is the gradient of polar angle, "$A = \nabla\theta$." That's "why" the scalar curl of $A$ is $0$ and $A$ is locally gradient. The snag is, a continuous choice (or branch) of polar angle function does not exist in $R$. Instead, polar angle is multi-valued, well-defined only up to added integer multiples of $2\pi$. If $D$ is an open subset of $R$ on which a branch of polar angle $\theta$ exists, then $A = \nabla\theta$ in $D$.
Physically, think of $\theta$ representing height in a parking garage (with infinitely many levels) sitting over $R$. Locally (i.e., over a disk contained in $R$) there is a continuous choice of height. If we follow a path enclosing the origin, however, we change levels. But because any two levels of the garage are parallel, the gradient of height is well-defined even though height itself is not.
If $A$ were gradient, incidentally, the M. C. Escher lithographs Ascending and Descending and Waterfall would not be paradoxical. Welcome to the wonders of "cohomology!"

Now to the question: If $(x_{0}, y_{0})$ is an arbitrary point of the plane, we may similarly analyze the "translated" vector fields
$$
A(x, y) = \frac{(-(y - y_{0}), x - x_{0})}{(x - x_{0})^{2} + (y - y_{0})^{2}},\qquad
B(x, y) = \frac{(x - x_{0}, y - y_{0})}{(x - x_{0})^{2} + (y - y_{0})^{2}}
$$
in the plane with $(x_{0}, y_{0})$ removed. The first is not gradient, i.e., not conservative, geometrically because there is no global branch of "polar angle from $(x_{0}, y_{0})$." The second field is gradient, exercise. (The second field is not relevant to what follows, but does relate to your linked question.)
Generally, a sum of gradient fields is gradient because

 If $F = \nabla f$ and $G = \nabla g$, then $F + G = \nabla f + \nabla g = \nabla(f + g)$,

and a sum of a gradient and a non-gradient field is non-gradient because

 Contrapositively, if $F$ and $F + G$ are gradient, so is $G = (F + G) - F$.

A sum of non-gradient fields can be either gradient or non-gradient, but loosely the sum can only be gradient if "the non-gradient parts cancel." This type of behavior has to be carefully arranged. For example, if $D$ is the complement of the interval $[-1, 1]$ on the $x$-axis (or is the annulus in the question), then the fields
$$
F_{1}(x, y) = \frac{(-y, x - 1)}{(x - 1)^{2} + y^{2}},\qquad
F_{2}(x, y) = \frac{(-y, x + 1)}{(x + 1)^{2} + y^{2}}
$$
are non-gradient in $D$ because

 A closed path in $D$ might enclose $(1, 0)$ and/or $(-1, 0)$.

Their sum is also non-gradient in $D$ because

 The added "$2\pi$-ambiguities" are summed if we traverse a loop enclosing the slit $[-1, 1]$.

Their difference $F_{2} - F_{1}$, on the other hand, turns out to be gradient in $D$, because

 The added "$2\pi$-ambiguities" cancel if we traverse an arbitrary closed loop in $D$. Geometrically, there is a branch of the difference of polar angle from $(1, 0)$ and polar angle from $(-1, 0)$.

Exercise: Is the difference $F_{2} - F_{1}$ conservative in the twice-punctured plane with $(\pm1, 0)$ removed?

 No. (!) A closed path in the twice-punctured plane can, for example, enclose $(1, 0)$ without enclosing $(-1, 0)$. The work done around such a path is $2\pi \neq 0$.

