Show function $f(x,y)=(x^2-y^2,2xy)$ is $1$-$1$ by Inverse Function Theorem

Solution 1:

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 $b=y$

Solution 2:

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.

Solution 3:

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 $y = \frac{t}{2x}$.