Showing that a multivariable function is one to one

Question

I am stuck with the following problem

I am given the function $f$ such that $f(x,y)=(x^2-y^2,2xy)$ I am supposed to show that the function is one to one.

For a function to be one to one, $f'>0$. If I differentiate this function, I get the matrix

$\begin{matrix} 2x&-2y\\2y&2x\end{matrix}$

How do I proceed from this step further?

Any hint would be appreciated

Thanks in advance!

Answer 1

Let $f(x,y)=(x^2-y^2,2xy)$ for $x,y>0$.

If $f(x,y)=f(a,b)$, then $x^2-y^2=a^2-b^2$ and $2xy=2ab$.

Then $x^2-a^2=y^2-b^2\implies (x-a)(x+a)=(y-b)(y+b)$ and $2xy=2ab\implies xy=ab$.

Assume that $x\ne a$; then without loss of generality, we can assume that $x>a$.

Then $xy=ab\implies y<b$, so $(x-a)(x+a)>0$ while $(y-b)(y+b)<0$; and this gives a contradiction.

Therefore $x=a$, and $xy=ab\implies y=b \implies (x,y)=(a,b)$.

Thus $f$ is 1-1.

Answer 2

It is one-to-one on positive numbers, but not on integer numbers, since $f(x,y)=f(-x,-y)$.

If we assume that we have to show that it is one-to-one on positive numbers, we need to find how to invert this function. Note that $(x^2-y^2)^2+(2xy)^2 = (x^2+y^2)^2$. Thus $$f^{-1}(u,v) = \left(\sqrt{0.5(\sqrt{u^2+v^2}+ u)}, \sqrt{0.5(\sqrt{u^2+v^2} - u)}\right)$$