I cannot grasp the concept of index of the singular point of vertor fields on the plane

Question

My book, Intuitive Topology by V.V.Parasolv, says that

 Let us go around a contour without self-intersections encircling exactly one singular 
point. For each point that we visit, consider the corresponding vector, attacing it to some
fixed point, possibly changing in length. 
 Let us see how many complete revolutions it performs while we go around our contour.
We count a full revolution with a "+" sign if it was performed in the same direction of 
rotation as our trip around the contour and with a "-" otherwise.
 The total number of signed revolutions is said to be the index of the singular point.

Then look at the vector field $a$, s.t $a(x,y)=(x,y)$

It is sure that $(0,0)$ is the only singular point in $a$

Isn't it total number of signed revolutions(i.e. index of the singular point) is infinite? I think $x^2+y^2=t^2$ for any $t \in R$ satisfies condition above. But It is 1. Why is it?

Answer 1

Along your curve, the vector always points away from the origin. As the vector follows the curve, it rotates once relative to the axes per revolution. Hence, its index is 1.

I think maybe what's confusing you is that the index is determined by any closed curve surrounding the one singular point. To measure the index, you go around one time on one closed curve.