Given a function $f:\mathbb{R}^2 \to \mathbb{R}^2$, to find its inverse near a given point

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a function given by $$f\left(x,y\right)=\left(x-y,xy\right),\,\, \left(x,y\right) \in \mathbb{R}^2$$ Question : What is the inverse of $f$ near the point $\left(2,-3\right)$?

Upon checking the conditions, inverse function theorem gives me the existence of $f^{-1}$, that $f^{-1}$ is $C^1$ under appropriate condition and an explicit form of the derivative of $f^{-1}$. However, I do not understand how to calculate an explicit form of $f^{-1}$ using inverse function theorem.

By direct calculation, I find : $$f^{-1}\left(u,v\right)=\left(\frac{2v}{-u\pm\sqrt{u^2+4v}},\frac{-u\pm\sqrt{u^2+4v}}{2}\right)$$ Two issues : $1$. It appears to be a one-to-two(!) map.

$2$. At the point $\left(2,-3\right)$, both the arguments are complex numbers!

Any help would be much appreciated.

Answer 1

We have that $f(2,-3)=(5,-6)$, hence $f^{-1}(5,-6)=(2,-3)$. Now compute $f^{-1}(5,-6)$ with

$f^{-1}\left(u,v\right)=\left(\frac{2v}{-u\pm\sqrt{u^2+4v}},\frac{u\pm\sqrt{u^2+4v}}{2}\right)$ and you will see which sign $ \pm$ is valid !

Answer 2

I think you made a small error. $f(2,-3)=(2-(-3),2\times(-3)) = (5,-6)$. $\sqrt{u^2 + 4v}\Big|_{(u,v) = (5,-6)} = \sqrt{25-24} = 1$. So the possible inverses for the point $(5,-6)$ are

$$f^{-1}(5,-6) =\left(\frac{-12}{-5\pm1},\frac{-5\pm1}{2}\right) $$ Since we want this to be $(2,-3)$, we are forced to pick the signs $$(2,-3) = f^{-1}(5,-6) =\left(\frac{-12}{-5-1},\frac{-5-1}{2}\right) $$

Answer 3

Attempt : $f^{-1}$ near $f(2,-3)$.

$a= x-y$; $b=xy;$

$a^2+4b =(x+y)^2;$

$x+y = -\sqrt{a^2+4b}$ (why minus ?)

$a=x-y$;

$2x= a-\sqrt{a^2+4b};$

$2y= -a-\sqrt{a^2+4b};$

Check:

$x-y=a$; $xy=-(1/4)(a^2-(a^2+4b))=b$.