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Consider the function $f:\mathbb{R^2}\to \mathbb{R^2}$ defined as $f(x,y)=(x^2-y^2,2xy)$ then the image of teh ball $B(P,\frac{1}{2})$, where $P=(0,1)$ is the centre and the radius is $\frac{1}{2}$ is,

A. a closed set

B. A open set

C. neither open nor closed set

D An unbounded set

I computed what this $B(p,\frac{1}{2})$ is, which is the interior of circle $(x-1)^2+y^2=\frac{1}{4}$.But how to obtain its image under the given $f(x,y)$.

Thanks in advance!!

math student
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2 Answers2

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It seems we can consider $f$ as a map from $\Bbb C$ to $\Bbb C$, which satisfies the Cauchy-Riemann equations.

So $f$ is analytic. In fact everywhere, so entire. Actually we have $f(z)=z^2.$

Two things:

  • by the open mapping theorem, $f$ is open
  • by Liouville $f$ is unbounded

Thus $b)$ appears correct. I don't see how to apply the second part.

calc ll
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  • May be I am mistaken, but $f$ is continuous as $\mathbb{R^2}\to \mathbb{R^2}$ and, as closed ball is compact, then $f$ is bounded on closed ball, so, on open ball also. No? – zkutch Feb 12 '23 at 11:35
  • Yes. But since $f$ is entire and not constant, it's unbounded. It's just an observation. @zkutch – calc ll Feb 12 '23 at 11:41
  • Of course $z^2$ is unbounded on $\Bbb C$, but OP is about image of ball. – zkutch Feb 12 '23 at 11:57
  • I know I just made the observation. The answer is $b).$ @zkutch – calc ll Feb 12 '23 at 11:58
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The answer is b. It cannot be d, since continuous functions send bounded sets into bounded sets (see here).

Moreover, $f$ is injective on $y>0$ (where our ball lies). To see this, let's see if $(u,v)=f(x,y)$ has a unique solution there. Since $y>0$ there holds $x=v/(2y)$. From $x^2-y^2=u$ we get $y^4+uy^2-v^2/4=0$. There a unique solution to this quadratic equation in $y^2$, as $y^2>0$. Thus, $f$ is injective in a neighborhood of your ball.

But continuous injective functions are open (this fact is not at all trivial: see here), and so, the image of our open ball is itself open.

Lilla
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