inverse of cardioid $(x,y) \rightarrow (x+0.5(x^2-y^2),y(1+x))$ $(x,y) \rightarrow (x+0.5(x^2-y^2),y(1+x))$ and $(x,y) \in \mathbb B_1(0) \subset \mathbb R^2$ the unit ball
How do I find the inverse of the given function ?
Would appreciate any help
Edit: I tried to find the inverse function as suggested in the comments by using the equation for the parabola and  points that lie on the inverse function.
Using the points (see the comment) I found $y=-0.5x^2$ which is obviously wrong.
Would appreciate it, if anyone could guide me through the calculation
 A: The inverse of a cardioid is a parabola. See, for example, the wikilink here.
Even without a formula to find the inverse directly, you can determine a parabola from three generic points (or from its vertex and a single other point). So, if you can find three points on the inverse's curve, then you should be able to write down the unique parabola containing those points; once you have done that, you can compose the cardioid function with the parabolic function to verify that the latter is, indeed, its inverse.
A: You want a function $g$ such that $f(g(x,y))=g(f(x,y))=(x,y)$ or $$g \left( x+\frac 12 (x^2-y^2), y(x+1) \right)= (x,y). $$
Let $a=x+\frac 12 (x^2-y^2), b=y(x+1)$ then you want to represent $x,y$ in terms of $a,b$. This is equivalent to solving system of equation 
$\begin{cases} a= \frac 12(x+1)^2-\frac 12y^2-\frac 12 \\ b=y(x+1) \end{cases}$ with respect to $x,y$. 
The answer I got is $x= -1\pm \sqrt{\frac{2a+1+\sqrt{(2a+1)^2+4b^2}}{2}},
y= \pm\sqrt{\frac{\sqrt{(2a+1)^2+4b^2}-2a-1}{2}}$. 
Hope this helps.
