I'm struggling to understand how to easily compute the index of a singular point of a general vector field.
For example with the following:
I have solutions which say "in each quadrant the angle turns by $\frac{3\pi}{2}$ so the total turning angle is $6\pi$, thus the index is 3". But I really don't see at all how the angle turns by $\frac{3\pi}{2}$ in each quadrant.
Visually, draw a small circle around the singular point (where small means enclosing only one singular point). Traverse the circle once counterclockwise, counting how many times the field rotates relative to any convenient constant field (e.g., the rightward-pointing field, in green). The signed number of turns (positive for counterclockwise, negative for clockwise) is the index.
The fields shown have index $3$ and $-3$, respectively.