$f(x,y)= (x^2-y^2, 2xy)$. Show $f$ is one-to-one Let $A:=\{(x,y)\in \mathbb{R},x>0\}$. Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined by $f(x,y)= (x^2-y^2, 2xy$). Show that $f$ is one-to-one on $A$.
My attempt:
$$f(x_1,y_1)=f(x_2,y_2)$$ so
$$\tag{$*$} x_1^2-y_1^2=x_2^2-y_2^2$$
and
$$2x_1y_1=2x_2y_2$$
$$\tag{$**$}x_1y_1=x_2y_2$$
From $(**)$, we have two cases. Either $y=0$ or $y\neq 0$. 
CASE $1$ : $y=0$. From $(*)$: $$x_1^2=x_2^2$$ Taking the square root of both sides $$x_1=x_2$$ where $x_1,x_2>0$. 
CASE $2$: $y \neq 0$ 
I am stuck on Case $2$...
 A: A slick way of approaching this is as follows. If $z =x+iy$, then 
$$ z^2 = (x^2-y^2) + 2xyi$$
so you function assigns a complex number $z = (x,y)$ to its square. But the square roots of $z\neq 0$ are two, and if $w$ is one, the other is $-w$. Now only one of $w$ has positive real part, so $z\longmapsto z^2$ is injective on $\Re(z) > 0$. 
A: $f(x,y)= (x^2-y^2, 2xy)
$.
If
$f(x, y) = f(u, v)$
then
$x^2-y^2 = u^2-v^2$
and
$2xy = 2uv$.
Squaring these,
$x^4-2x^2y^2 + y^4 = u^4-2u^2v^2+v^4$
and
$4x^2y^2 = 2u^2 v^2$,
Adding these,
$x^4+2x^2y^2 + y^4 = u^4+2u^2v^2+v^4$
or
$(x^2+y^2)^2 = (u^2+v^2)^2$.
Since both $x^2+y^2$
and
$u^2+v^2$
are positive,
we can take the square root
to get
$x^2+y^2 = u^2+v^2$.
Since
$x^2-y^2 = u^2-v^2$,
adding these we get
$2x^2 = 2u^2$;
subtracting them we get
$2y^2 = 2v^2$.
Therefore
$x^2 = u^2$
and
$y^2 = v^2$.
Therefore $x = \pm u$
and
$y = \pm v$.
By assumption
(on $A$),
$x \ge 0 $ and $u \ge 0$.
Therefore
$x = u$.
Since
$2xy = 2uv$,
$y = v$.
If you allow $x < 0$,
this is equivalent to
the two-valued nature of
the complex square root.
