Show function $f(x,y)=(x^2-y^2,2xy)$ is $1$-$1$ by Inverse Function Theorem I'm trying to prove the problem below - which comes from Munkres' "Analysis on Manifolds" book in the section on the Inverse function theorem.
Since its in the chapter on the Inverse Function Theorem I figured I'd start by showing that $f$ satisfies the conditions of the theorem.  Writing the Jacobian shows that it's both $C^r$ and we get $\det f'(x,y)=4x^2+4y^2\neq0$ when $x>0$.  So we can apply the theorem but I'm unsure of how to proceed to show that $f$ is $1$-$1$.  And I didn't see how to use the hint they provided.
Thanks!


Let $f\colon \mathbf R^2\to \mathbf R^2$ be defined by the equation
  $$f(x,y)=(x^2-y^2,2xy).$$
  (a) Show that $f$ is one-to-one on the set of all $(x,y)$ with $x>0$. [Hint: If $f(x,y)=f(a,b)$, then $\|f(x,y)\|=\|f(a,b)\|$.]

 A: Compute the norm: $$\|f(x,y)\|=\sqrt{(x^2-y^2)^2+4(xy)^2}=\sqrt{x^4-2(xy)^2+y^4+4(xy)^2}=x^2+y^2$$
so you get: $f(x,y)=f(a,b)\implies (x^2-y^2,2xy,x^2+y^2)=(a^2-b^2,2ab,a^2+b^2)$, use the first and last coordinates to show that $(a,b)=(\pm |x|,\pm |y|)$ (add and subtract them), use $A$ to conclude that $a=x$ ($x,a>0$) and use the second coordinate to get $b=y$
A: This is simpler in polar coordinates: The map is
$$r(\cos t,\sin t) \to r^2(\cos^2t - \sin^2t, 2\cos t \sin t) = r^2(\cos 2t,\sin 2t).$$
Thus the ray in the right half plane making angle $t$ with the $x$-axis is sent injectively to the ray making angle $2t$ with the $x$-axis. Overalll injectivity follows.
A: We can prove $f$ is 1-1 by showing there is an explicit inverse function:
$f(x,y) = (x^2-y^2, 2xy) = (s.t)$
$\Rightarrow s^2 + t^2 = x^4 + 2 x^2y^2 + y^4 = (x^2 + y^2)^2$
$\Rightarrow x^2 = \frac{s + \sqrt{s^2 + t^2}}{2}$
There is no ambiguity over the sign of $\sqrt{s^2+t^2}$ here because we know $x^2$ must be $\ge 0$. Since $x > 0$ we have $x = +\sqrt{x^2}$ and then $y = \frac{t}{2x}$.
