$f(x,y)=(x^2-y^2,2xy)$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)$ Let $f:\mathbb R^2\to\mathbb R^2$ be defined by the equation 

$$f(x,y)=(x^2-y^2,2xy).$$

Show that $f$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$.
 What is the set $f(A)?$

The inverse function theorem does not seem to help since we are talking about global invertability. To show injective assume that 
$$x^2-y^2=w^2-z^2$$ and 
$$2xy=2wz$$
We then have on squaring and adding,
$$(x^2+y^2)^2=(w^2+z^2)^2$$ this implies that
$$x^2+y^2=w^2+z^2$$ from which we get $x^2=w^2$ and since we assume that these are positive we have $x=w>0$ and now the second equation gives $y=z$ proving injectivity.
To determine the image 
set $$a=x^2-y^2$$
$$b=2xy$$ then we see that 
$$x=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$$
Now the condition $x=0$ gives
iff $a+\sqrt{a^2+b^2}=0$ that is $b=0$ and $a \leq 0$, so the negative $x$ axis is not in the image, all other points are. 
