# Showing a function is 1-1?

## Homework Statement

Show that the function f(x,y) = (x² - y², 2xy) is 1-1 on the the set A where x > 0.

## Homework Equations

The problem gives a hint: If f(x,y) = f(a,b), then ||f(x,y)|| = ||f(a,b)||, where || || is the euclidean norm on R².

## The Attempt at a Solution

I suppose this shouldn't be hard, but I don't know how to do it. So showing that f(x,y) is 1-1 means that if f(x,y) = f(a,b) then (x,y) = (a,b). So using the hint I have:

||(x² - y², 2xy)|| = ||(a² - b², 2ab)||

Expanding I have:

(x² - y²)² + (2xy)² = (a² - b²)² + (2ab²); I'll ignore the square roots

But here I've tried factoring it in tons of different ways to show that x = a (and somehow using that x, a > 0) and y = b. But I don't see it. The next problem says that the hint is also once again useful, so could maybe someone help me out? Oh and please don't use complex analysis as I do know this is z², but I'd like to do it using the hint. I'm studying for a term test for tomorrow.

Thanks!!

Another hint would probably be enough...just give me something! lol

Last edited:

Anyone? I know almost everyone could do a question like this on here. So please some help?

LCKurtz
Homework Helper
Gold Member
Well, I dunno about the hint but what about trying something like this.

Suppse a > b. Then (x,y) lies on the right branch of the hyperbola x2 - y2 = a2 - b2.

(x,y) must also lie on the orthogonal hyperbola xy = ab. This hyperbola lies in quadrants 1 and 3 or 2 and 4 depending on the sign of ab. Either way it only can intersect the first hyperbola in one point. Since (x,y) and (a,b) lie on the intersection, of which there is only one, (x,y) = (a,b).

The other cases should be similar.

Hmm that doesn't seem over complicated does it? The hint seems to lead to the following:

(x² + y²)² = (a² + b²)²

But would this imply anything? I don't seem to be able to get that x = a and b = y?

Waiiitt a second...so we have:

(x² + y²)² = (a² + b²); positive square root since both sides are sum of squares

Basically we have then:

x² + y² = a² + b²
(x² - a²) + (y² - b²) = 0
||(x - a, y - b)|| = 0

However, by properties of the norm this is only true when (x - a, y - b) = 0, ie x = a and y = b.

But how does x > 0 come into play here?

Btw, is this right??

LCKurtz
Homework Helper
Gold Member
Waiiitt a second...so we have:

(x² + y²)² = (a² + b²); positive square root since both sides are sum of squares

Basically we have then:

x² + y² = a² + b²
(x² - a²) + (y² - b²) = 0
||(x - a, y - b)|| = 0

||(x,y) - (a,b)||2 = (x-a)2 + (y-b)2, not (x² - a²) + (y² - b²)

Ahh yes. So I guess it works out. Thanks for your help LCKurtz! :)