I am looking at theory about vector fields in the plane and about their indices, I have found classic examples but I was wondering if anyone knows a plane vector field that behaves as follows:
Considering that the lines in the image start the plane in equal parts, I have managed to obtain that the index is $-2$, but I do not know which face this vector field has. If anyone has a reference to where I can find examples of vector fields that have indices $+2$, $+3$ or $-3$, $-4$, I would ask you to mention a reference. Thanks in advance.
Rather than writing down a formula, let me tell you how to write it down for yourself. The resulting vector field, with a singularity at the origin of index $1-n$, will be denoted $\vec v_n(x,y)$.
Here's the clue for deriving a formula for $\vec v_n(x,y)$: in complex coordinates, the $n$ inward rays pass through the $n^{\text{th}}$ roots of unity. I presume that you already have in your hands a formula for $\vec v_2(x,y)$, whose $2$ inward rays pass through the square roots of unity.
Consider the "function" of a complex variable given by $w = F(z) = z^{2/n}$; in fact one can choose any branch of this function defined on the complement of any ray, although as usual you'll need at least two branches to cover the whole complex plane.
Write this function in real coordinates $w = u+iv$, $z=x+iy$, obtaining $$(u,v) = F(x,y) = (f(x,y),g(x,y)) $$ Compute its total derivative $D_{(x,y)} F$.
So, starting from a formula for $\vec v_2(x,y)$, for $(u,v) = F(x,y)$ we have $$\vec v_n(u,v) = D_{(x,y)} F(\vec v_2(x,y)) $$ Again, I'll emphasize that one needs to use two well-defined functions $F$ to verify this computation, but in the end one will get a well-defined formula expressed in $(u,v)$ coordinates. A key feature of this computation is that the vector field, while continuous at $(0,0)$, is not smooth at $(0,0)$.