Showing that a transformation maps the exterior of a disk onto the exterior of an ellipse I'm given a transformation T:
$u(x,y)=2x+\frac{2x}{x^2+y^2}$,  $v(x,y)=2y-\frac{2y}{x^2+y^2}$
And I'm asked to show that the transformation maps the exterior of the disk of radius 2 $D:=\{(x,y)|x^2+y^2\geq4\}$ onto the exterior of the ellipse $\{(u,v)|u^2/25+v^2/9\geq1\}$.
My first attempt (which I thought was all I needed to do) was to simply put $u^2$ and $v^2$ onto the ellipse to get the disk, but that did not work.
My question is: what am I missing? because I though the problem was pretty straightforward.
After that I need to compute the Jacobian and use it to calculate the area between two ellipses, but again, the problem seems a little odd since the transformation should make the integral easier to compute but I computed the Jacobian and it does not seems like it is helping.
I'm not asking for the solution, I just want to know why my attempt did not work and I would appreciate any hints or explanations about how to solve it, or any insights about transformations that maybe I'm not taking into account!
 A: Under the assumption that $x^2+y^2=4$, you should have been able to prove that $$\frac{1}{25}\left(2x+\frac{2x}{x^2+y^2}\right)^2 + \frac{1}{9}\left(2y-\frac{2y}{x^2+y^2} \right)^2 = 1,$$so possibly you made an algebra mistake. But this would have only showed that $T$ takes the circle into the ellipse. Note that $$(\Bbb R^2 \setminus \{(0,0)\}) \setminus \{(x,y): x^2+y^2 = 4\} \quad \mbox{and} \quad (\Bbb R^2\setminus \{(0,0)\}) \setminus \{ (u,v) : u^2/25 + v^2/9 =1\}$$each have two connected components, so $T$ has to map connected components to connected components, by continuity. Then to check that $D$ gets mapped to the exterior of the ellipse (as opposed to the punctured interior), it would suffice to test one point: for example, $T(3,0) = (20/3,0)$.
In any case, Wolfram Alpha gives that the expression expands to $$\frac{8 x^2}{25 (x^2 + y^2)} + \frac{4 x^2}{25 (x^2 + y^2)^2} - \frac{8 y^2}{9 (x^2 + y^2)} + \frac{4 y^2}{9 (x^2 + y^2)^2} + \frac{4 x^2}{25} + \frac{4 y^2}{9},$$so with $x^2+y^2=4$ this indeed gives $1$.
I'd like to offer another point of view, though, simpler and more enlightening: it might be interesting to set $\Bbb R^2 \cong \Bbb C$, so that $$T(z) = 2\left(z + \frac{1}{z}\right).$$If $z(t) = 2{\rm e}^{{\rm i}t}$ parametrizes the circle, then we have $$T(z(t)) = 2\left(2{\rm e}^{{\rm i}t} + \frac{1}{2}{\rm e}^{-{\rm i}t}\right) = 4{\rm e}^{{\rm i}t} + {\rm e}^{-{\rm i}t} = 5\cos t + 3{\rm i} \sin t.$$This is explictly a parametrization of the ellipse, in view of ${\rm e}^{{\rm i}t} = \cos t + {\rm i}\sin t$.
