Let $f : \mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation $f(x,y)=(x^2-y^2, 2xy)$. If $g$ is the inverse function, find $Dg(0, 1)$.
I know that $Dg(y)=[Df(g(y))]^{-1}$, with which $Dg(0,1)=[Df(g(0,1))]^{-1}$ and as $Df(x,y)=\begin{bmatrix}2x & 2y\\2y &2x \end{bmatrix}$, I just need to know how much $g(0,1)$ is, but I do not know how to find it, could someone help me please? How can I find an explicit formula for $g(x,y)$? Do I have to do this or is there another way to do it? Thank you very much.
Edit: This is a question from book "Analysis on manifolds" by Munkres, and the complete exercise is:
Let $f : \mathbb{R}^2\to \mathbb{R}^2$ be defined by the equation $f(x,y)=(x^2-y^2, 2xy)$.
(a) Show that $f$ is one-to-one on the set $A$ consisting of all $(x,y)$ with $x > 0$. [Hint: If $f(x,y)= f(a,b)$, then $||f(x, y)|| = ||f(a,b)||$]
(b) What is the set $B = f(A)$?
(c) If $g$ is the inverse function, find $Dg(0, 1)$.
I already know how to solve (a) and (b), I am stuck solving (c), could someone help me please? Thank you.
Note that $f$ is the "real expression" of the analytic function $\tilde f(z):=z^2$. There are two points $z$ with $\tilde f(z)=i$, namely the points $\pm {1\over\sqrt{2}}(1+i)$. Correspondingly we have two points $(x_j,y_j)$ with $f(x_j,y_j)=(0,1)$, namely the points $\pm\bigl({1\over\sqrt{2}},{1\over\sqrt{2}}\bigr)$. At each of these points the derivative $Df(x_j,y_j)$ has nonzero determinant (check this!); therefore $f$ maps suitable neighborhoods $U_j$ of these points diffeomorphically onto a neighborhood $V$ of $(0,1)$. As a consequence we have two local inverse functions $$g_j:\quad V\to U_j\ ,$$ each of them with its own derivative at $Dg_j(0,1)$, which is the inverse of $Df(x_j,y_j)$.