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I was given this exercise:

Let $U=\{(x,y): 1 and $f:U\rightarrow \mathbb {R^2}$ defined by:

$f(x,y)=\left(\frac {x^2-y^2}{r},\frac {2xy}{r}\right)$

where $r=\sqrt {x^2+y^2}$.

Then I have to show that $f(U)=U$ but $f$ is not injective... Think that polar coordinates might help.. but how? I'm kind stuck.

Thanks for any help!

4 Answers 4

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$f$ is not injective because $f(-x,-y)=f(x,y)$, so for example the two points $(-1,-1)$ and $(1,1)$ are mapped to the same point $(0,\sqrt2)$. Polar coordinates are indeed a good start. Write $x=r \cos(\phi)$ $y=r \sin(\phi)$ Then $f(x,y)=\left(\frac {r^2\cos^2(\phi)-r^2\sin^2(\phi)}{r},\frac {2r \cos(\phi)r \sin(\phi)}{r}\right)=\left(r(\cos^2(\phi)-r^2\sin^2(\phi)),r(2\sin(\phi)\cos\phi))\right)=\left(r\cos(2\phi),r\sin(2\phi)\right)$ So, $f$ sends $(r,\phi)$ to $(r,2\phi)$. Since $U$ is just the annulus $1 and $r$ is left invariant under $f$, $f$ sends $U$ to $U$ and every $(r,\phi) \in U$ has exactly two preimages $(r,\phi/2)$ and $(r,\phi/2+180°)$

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    @Melvin See my answer below.2012-11-26
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With polar coordinates:

$x=r\cos t\,\,,\,\,y=r\sin t\,\,,\,\,r\geq 0\,\,,\,0\leq t\leq 2\pi\Longrightarrow$

$U=\{(r,t)\;;\;\;1

Thus, $\,f(r,t)=f(r,t+\pi)\,$

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Use complex numbers!

Let $z = x+iy$, then $|z| = \sqrt{x^2+y^2}$ and $z^2 = (x^2-y^2)+2ixy$. Thus, the complex function: $f : z \mapsto z^2/|z|$ can be identified with your function by taking the real part as the first coordinate, and the imaginary part as the second coordinate.

You hav $U = \{z \in \mathbb{C} : 1 < |z|^2 < 2 \}.$ Let $z \in U$, and consider

$|f(z)| = \left|\frac{z^2}{|z|}\right| = \frac{|z|^2}{|z|} = |z|$

It follows that $|z| = |f(z)|.$ Thus, $z \in U \iff f(z) \in U.$ Consider the two facts: $z \in U \implies f(z) \in U$ tells us that $f(U) \subseteq U,$ while $f(z) \in U \implies z \in U$ tells us that $U \subseteq f(U).$ It follows that $U = f(U).$

The function is clearly not injective since $f(\pm 1) = 1.$ In my notation $1 = 1 + 0i \in \mathbb{C},$ which corresponds to $(1,0) \in \mathbb{R}^2$ in your notation.

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Hint:

Given some $(x',y')$ s.t $1 solve $x'=\frac{x^{2}-y^{2}}{r},y'=\frac{2xy}{r}$ for $x,y$ and show that $1.

That proves that $U\subseteq f(U)$.

To prove $f(U)\subseteq U$ that $x,y$ s.t $1 and prove that $1<\left(\frac{x^{2}-y^{2}}{r}\right)^{2}+\left(\frac{2xy}{r}\right)^{2}<2$

Both are done with relatively simple algebra.