Is $f(x,y) = ( \frac{x}{x^2+y^2},\frac{2y}{x^2+y^2} )$ injective?

Question

I came across this function in a context in which I need to know if it is injective. The function is $f:\mathbb{R}^2\backslash\{(0,0)\} \longrightarrow \mathbb{R}^2\backslash\{(0,0)\}$ and defined by $$f(x,y) = \Biggl( \frac{x}{x^2+y^2},\frac{2y}{x^2+y^2} \Biggl).$$

I guess it is injective. I have failed to find counterexamples, but I couldn't prove that it is injective. Maybe I should use some result, but I couldn't find any that seemed useful.

Hints or answers are much appreciated.

Change the $2y$ to $-y$ (which has no effect whether it is injective) and you have the map $z \mapsto 1/z$ on $\mathbb C \setminus {0}$. — Erick Wong, Apr 02 '13 at 21:16
Thanks for all the answers and comments, they've helped my understanding of the subject. — Wheepy, Apr 02 '13 at 22:32

score 6 · Accepted Answer · answered Apr 02 '13 at 21:15

6

Try solving $(u,v) = f(x,y)$. Here is a hint to start: Let $s = x^2+y^2$, this gives $u = \frac{x}{s}$, $\frac{v}{2} = \frac{y}{s}$. Squaring gives $u^2+(\frac{v}{2})^2 = \frac{1}{s}$, from which we get $s = \frac{1}{u^2+(\frac{v}{2})^2}$. Now figure out what $x,y$ are.

Then $x = s u$, $y = s \frac{v}{2}$ give $x = \frac{u}{u^2+(\frac{v}{2})^2}$, $y = \frac{\frac{v}{2}}{u^2+(\frac{v}{2})^2}$.

Since $f$ is invertible, it follows that it is injective.

answered Apr 02 '13 at 21:15

copper.hat

172,524

How did you make the invisible box? – Berci Apr 02 '13 at 21:23
3

I used >! at the start. (You can edit my answer to look.) – copper.hat Apr 02 '13 at 21:24
Ah, indeed. Thanks. – Berci Apr 02 '13 at 22:11

score 3 · Answer 2 · answered Apr 02 '13 at 21:10

3

Hint Express $x$ and $y$ from $u=\frac{x}{x^2+y^2}$ and $v=\frac{2y}{x^2+y^2}$.

One more hint: for start, calculate $u^2+(v/2)^2$.

answered Apr 02 '13 at 21:10

Berci

90,745

Is $f(x,y) = ( \frac{x}{x^2+y^2},\frac{2y}{x^2+y^2} )$ injective?

2 Answers2