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

Question

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 $\|f(x,y)\|=\|f(a,b)\|$.]

It is straightforward to compute an explicit inverse, but an answer based on the hint is not leaping out at me. — copper.hat, Jul 01 '18 at 23:33
@Holo Also, I find that it's more convenient to write $x^2$ than $x²$ , as you did in your comment. — Ben Grossmann, Jul 02 '18 at 00:01
@Omnomnomnom I posted it as answer. Also, normally write x^2 but right now I'm in bed with my phone, so the autocorrect change my x to X all the time if I use ^. — ℋolo, Jul 02 '18 at 00:04
@Holo Thanks for using your phone to help me out instead of watching world cup highlights! — john fowles, Jul 02 '18 at 00:12
@johnfowles don't mind that, sport was never my interest. I would have used the computer instead but my laziness took over and I don't want to get up( we all have those days) — ℋolo, Jul 02 '18 at 00:15
Using Approach0 you can find some posts about the same function (or even about the same problem). For example: $f(x,y)=(x^2-y^2,2xy)$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)$ or Question from Munkres Analysis on Manifolds Inverse Function Theorem Section. (The latter is about the third part of the exercise. — Martin Sleziak, Jul 02 '18 at 04:49

score 5 · Accepted Answer · edited Jul 02 '18 at 04:51

5

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 $b=y$

edited Jul 02 '18 at 04:51

Martin Sleziak

53,687

answered Jul 02 '18 at 00:02

ℋolo

10,006

If $x >0, $ how does that force $x = a$? Can't we have $x = -a$ with $a < 0?$ So its not one to one? – Lemon Feb 22 '20 at 09:07
@Hawk Because $f(x,y)=f(a,b)\implies(x,y)=(a,b)$ in the set $A$ with the $x$-coordinate being positive so $x,a>0$ – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Feb 22 '20 at 09:59
@TheSimpliFire "$\implies (x,y) = (a,b)$" is what you need to show, you can't assume this. But I think I see where this is going. Since $x >0$ this also forces the $(a,b)$ to be in $x>0$, and hence $a >0$ so the case $-a > 0$ is not considered – Lemon Feb 22 '20 at 22:20
@Hawk That's what I was trying to say – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Feb 23 '20 at 09:01

score 5 · Answer 2 · answered Jul 02 '18 at 05:43

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.

score 2 · Answer 3 · answered Jul 02 '18 at 11:42

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

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

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