Find affine function to approximate local inverse Let
$$
f(x,y) = (x^2 - y^2, 2xy)
$$
then by the inverse function theorem $f$ is invertible locally at any point $(x,y) \neq (0,0)$ because
$$
\det Df(x,y) = \det \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix} = 4(x^2+y^2) > 0 \iff (x,y) \neq (0,0)
$$
it's also not globally invertible given the counterexample
$$
f(1,1) = (0,2) = f(-1,-1)
$$
But how would I find a linear affine function that approximates $f^{-1}$ near $f(1,-1)$?
 A: As a function $f:\Bbb{C}\to\Bbb{C}$ it is imply
$$f(z)=z^2$$
with some algebra we can find an inverse
$$f^{-1}(z)=\sqrt{z} = r^{\frac{1}{2}}e^{\frac{i\theta}{2}} = r^{\frac{1}{2}}\left[\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\right]$$
And using the half angle identities gives us that the inverses are of the form
$$f^{-1}(x,y) = \left(\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},\pm\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)$$
which is four possible options. Since we want a solution around $(1,-1)$, we can narrow this down to just
$$f^{-1}(x,y) = \left(\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}},-\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}\right)$$
$f(1,-1) = (0,-2)$, which is where we center our linear approximation like so:
$$\frac{d}{dz}\sqrt{z} = \frac{1}{2\sqrt{z}} = \frac{1}{2(1-i)} = \frac{1+i}{4}$$
which means as a function $f:\Bbb{R}^2\to\Bbb{R}^2$
$$f^{-1}(x,y) \approx (1,-1) + \begin{pmatrix}\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4}\end{pmatrix}\left(x,y+2\right)$$
