Find the image of a ring I'm working on the following problem:
Find the image of the ring defined by $4 \lt x^2 + y^2 \lt 16 $ under the mapping 
$$F(x,y) = \left(\frac{x}{x^2+y^2} , \frac{y}{x^2+y^2}\right)$$
It looks to me like the easiest way to do this would be to use polar coordinates, but I'm having some issues putting it all together. Any help would be much appreciated.
 A: Perhaps the easiest way to think of it is, as you suggest, polar coordinates.
In this case, your region is described by $2<r<4$. Further, the effect of $F$ on the polar coordinates is given by
$$
F:(r,\theta)\mapsto\left(\frac{r\cos\theta}{r^2},\frac{r\sin\theta}{r^2}\right)=\left(\frac{\cos\theta}{r},\frac{\sin\theta}{r}\right).
$$
Notice that if we consider all points with a fixed $r$ value, the image is simply the circle of radius $\frac{1}{r}$: all points clearly have this radius, and we get all of them because of the way that cosine and sine are included. Since we map all points with $2<r<4$, this implies that the resulting image is simply the set of all points whose polar coordinates satisfy $\frac{1}{4}<r<\frac{1}{2}$. 
A: Polar coordinates seem like a good idea here. In polar coordinates your ring is: $2 < r < 4$.
Note that $x = r\cos(\theta)$ and $y = r\sin(\theta)$, so then 
$$
\frac{x}{x^2 + y^2} = \frac{r\cos(\theta)}{r^2} = \frac{\cos\theta}{r}\\
\frac{y}{x^2 + y^2} = \frac{r\sin(\theta)}{r^2} = \frac{\sin\theta}{r}.
$$
Now think about fixing $r$, then for that $r$ you get all points
$$
\frac{1}{r}(\cos(\theta), \sin(\theta))
$$
where $\theta$ varies. What does this give you? What do you get when you then vary $r$ from $2$ to $4$?
