$f(x,y)=(x^2-y^2,2xy)$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)$

Question

Let $f:\mathbb R^2\to\mathbb R^2$ be defined by the equation

$$f(x,y)=(x^2-y^2,2xy).$$

Show that $f$ is one to one on the set $A$ consisting of all $(x,y)$ with $x>0$. What is the set $f(A)?$

Answer 1

I think this problem is pretty straightforward if we observe that the map

$f(x, y) = (x^2 - y^2, 2xy) \tag{1}$

may also be represented in terms of the complex variable $z = x + iy$ as

$f(z) = f(x + iy) = (x + iy)^2 = (x + iy)(x + iy) = (x^2 - y^2) + 2ixy = z^2, \tag{2}$

or more concisely

$f(z) = z^2. \tag{3}$

In terms of the polar representation $z = re^{i\theta} = r\cos \theta + ir\sin \theta$, (3) becomes

$f(re^{i\theta}) = r^2e^{2i\theta}, \tag{4}$

where, since $x > 0$, $\theta$ is restricted to $\theta \in (-\pi / 2, \pi / 2)$; we note that $r > 0$ as well. To see that $f(z) = f(re^{i\theta})$ is injective, suppose that

$f(z_1) = f(r_1e^{i\theta_1}) = f(r_2e^{i\theta_2}) = f(z_2) \tag{5}$

for some $z_1$, $z_2$, i.e. for some $r_1, r_2 > 0$ and $\theta_1, \theta_2 \in (-\pi / 2, \pi / 2)$. Then from (3)-(5) we have

$r_1^2e^{2i\theta_1} = r_2^2e^{2i\theta_2}, \tag{6}$

whence, taking absolute values,

$r_1^2 = r_1^2 \vert e^{2i\theta_1} \vert = \vert r_1^2e^{2i\theta_1} \vert = \vert r_2^2e^{2i\theta_2} \vert = r_2^2 \vert e^{2i\theta_2} \vert = r_2^2 \tag{7}$

since $\vert e^{2i\theta_i} \vert = 1$, $i = 1, 2$; and since the $r_i > 0$ we thus infer

$r_1 = r_2, \tag{8}$

and thus (6) yields

$e^{2i\theta_1} = e^{2i\theta_2} \tag{9}$

or

$e^{2i(\theta_1 - \theta_2)} = 1. \tag{10}$

(10) implies that

$2(\theta_1 - \theta_2) = 2n\pi \tag{11}$

for some integer $n$, whence

$\theta_1 - \theta_2 = n\pi; \tag{12}$

but the only way (12) can hold with $\theta_1, \theta_2 \in (-\pi / 2, \pi / 2)$ is if $n = 0$; thus

$\theta_1 = \theta_2, \tag{13}$

and so

$z_1 = r_1e^{i\theta_1} = r_2e^{i\theta_2} = z_2; \tag{14}$

$f(z)$ is injective. Examining $f(re^{i\theta}) = r^2e^{2i\theta}$ for $\theta \in (-\pi / 2, \pi / 2)$ and $r > 0$ we see that $r^2 > 0$ and $-\pi < 2\theta < \pi$; $f(A)$ is thus the complex plane $\Bbb C$ with the non-positive reals ommited, i.e $f(A) = \Bbb C \setminus \{x + iy \mid x \le 0, y = 0\}$. In real terms, this may be expressed as $f(A) = \Bbb R^2 \setminus \{(x, 0) \mid x \le 0 \}$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

Reading the comments, seems that the only thing that's left to do is describe the set $f(A)$. Look at the part of circles, that are contained in $A$, that is, consider the curves $$\gamma (t) = (r \cos t, r \sin t)$$ with $r > 0$ fixed and $\frac{- \pi}{2} < t < \frac{\pi}{2}$. This assures that $x > 0$. Now, we have: $$f(\gamma (t)) = (r^2 \cos 2t, r^2 \sin 2t)$$ So, $f$ takes a point of the plane that is in the circle of radius $r$, and sends it to the circle of radius $r^2$, after applying a rotation. Look at what happens to the points dangerously close to the $x$-axis. It's the most extreme situation. Now it's easy to see that $$f(A) = \Bbb R^2 \setminus \{ (x,y) \in \Bbb R^2 : x < 0 ~ \mbox{and}~ y = 0 \}$$

Answer 3

Here is an approach (which agrees with Ivo's answer). Let

$$ u=x^2-y^2,\quad v=2xy \implies u^2+v^2 = (x^2+y^2)^2. $$

If you consider any circle in the $xy$-plane centered at the origin with radius $r$ and $x>0$ we will have the corresponding image of the circle

$$ u^2+v^2 = r^4. $$

Note: (If you are familiar with conformal maps (complex variables) ) your map is nothing but the conformal map

$$ f(z)=z^2,\quad z=x+iy. $$

Answer 4

The inverse function theorem does not seem to help since we are talking about global invertability. To show injective assume that

$$x^2-y^2=w^2-z^2$$ and

$$2xy=2wz$$

We then have on squaring and adding,

$$(x^2+y^2)^2=(w^2+z^2)^2$$ this implies that
$$x^2+y^2=w^2+z^2$$ from which we get $x^2=w^2$ and since we assume that these are positive we have $x=w>0$ and now the second equation gives $y=z$ proving injectivity.

To determine the image set $$a=x^2-y^2$$ $$b=2xy$$ then we see that

$$x=\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$$ Now the condition $x=0$ gives iff $a+\sqrt{a^2+b^2}=0$ that is $b=0$ and $a \leq 0$, so the negative $x$ axis is not in the image, all other points are.