Eliminate the parameter and images Given $h : \mathbb{R}^2\to \mathbb{R}^2$ $$h(u,v) = (u^2 - v^2, 2uv)$$
where $A = \{(u,v) \in \mathbb{R}^2 : 0 < u < \infty,-\infty < v < \infty \} $. What is the set $h(A)$?
Here is what I did:
Let $x=u^2 - v^2$ and $y = 2uv$, square and add, i.e.
$x^2 + y^2 = (u^2 -v^2)^2+4u^2v^2= u^4 + v^4 -2u^2v^2+4u^2v^2 = u^4 + v^4 + 2u^2v^2 = (u^2 + v^2)^2$
It would seem to suggest I am getting a circle, but with a hole because of $u \neq 0$
 A: Note that $(u+i v)^2 = (u^2-v^2) + i(2uv)$,
so you are looking at squaring the
complex values in the first and fourth quadrants in the complex plane.
This would double all the args of each point.
The points in the first quadrant
with args from $0$ to $\pi/2$ would map into the points
with args $0$ to $\pi$,
or the upper half plane.
The points in the fourth quadrant
with args from $3\pi/2$ to $2\pi$ would map into the points
with args $3\pi$ to $4\pi$,
or the lower half plane.
Together, these fill the whole plane,
so $h(A)$ is all of $\mathbb{R}^2$.
A: The given map $h$ is the composition of two maps:
$$g(u,v) = (u^2 - v^2, 2 u v, u^2 + v^2)$$
and 
$$p(x,y,z) = (x, y, 0)$$
The first of these, $g$, can be implicitized to give $x^2 + y^2 = z^2$, which means that the range of this mapping lies on a cone whose axis runs in the $z$ direction. Here's a picture of a portion of this cone:

The range of the mapping $g$ does not cover the entire cone. Points $(x,0)$ with $x \le 0$ are excluded for reasons explained in my other answer. This is indicated by the (artificially enlarged) crack in the cone.
The second map, $p$ just projects this cone down onto the $xy$ plane. 
A: Another approach, for folks who like algebra better than geometry:
Given any $(x,y) \in \mathbb{R}^2$, we can find find polar coordinates $r$ and $\theta$ such that $x=r\cos\theta$ and $y=r\sin\theta$. Then set 
$$ u = \sqrt r \cos\tfrac{1}{2}\theta  \quad ; \quad v = \sqrt r \sin\tfrac{1}{2}\theta  $$
It is easy to verify that $u^2 - v^2 = x$ and $2uv = y$.
So, for any given $(x,y) \in \mathbb{R}^2$, we can find $u,v$ such that $h(u,v)=(x,y)$. This shows that $h(A)$ is (almost) all of $\mathbb{R}^2$. The only problems occur on the negative $x$-axis. If $y=0$, then  $uv=0$. But $u>0$, so this means $v=0$, so $x = u^2$, which must be positive. So, in fact, $h(A)$ is $\mathbb{R}^2$ minus the points $(x,0)$ where $x \le 0$.
