# Finding, writing down and visualizing images of functions

I apologize in advance if this is a duplicate (I would expect to see something like this on here), but I haven't found anything that matches my question.

I have this general difficulty (perhaps, to lack of practical experience) of finding and writing down images of functions given a domain. For example, I have recently seen this problem in a book:

Given $f(x,y) = \left(\begin{array}{c}u(x,y)\\v(x,y)\end{array}\right) = \left(\begin{array}{c}x^2-y^2\\2xy\end{array}\right)$, find image of $f$ when $x>0$, $y>0$, $x^2+y^2<1$

Basically, it's asking for a set of inequalities for what $u$ and $v$ trace out in the $u$-$v$ plane. How do I go about solving such problems for a general case of $f : \mathbb{R}^m \to \mathbb{R}^n$? For example, in the above problem, the domain of $f$ is a quarter of a circle of radius 1 in the first quadrant of the $x$-$y$ plane. How do I find what $f$ maps this to in the $u$-$v$ plane? Where do I start?

Update: So apparently (by empirically constructing a table of values), $f$ maps the quarter of a circle to a half-circle ($v$>0, $u^2+v^2<1$). Is there a way to see this intuitively? Most importantly, how can I get these two inequalities algebraically?

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In this particular case $f$ can be written as the complex function $f(z) = z^2$, and then working in polar coordinates it should be pretty clear what its image is. In general, I would start by writing down the image of the boundary, if there happens to be one, and then seeing if you can fill in everything else. I would also advise cutting up the domain into pieces (say based on a certain coordinate) and seeing if you can find the image of the pieces. – Qiaochu Yuan May 20 '11 at 15:59
Thanks. Is there a way to find the image analytically/algebraically though (see updated question)? – Phonon May 20 '11 at 16:40

If you want an algebraic solution, follow Qiaochu Yuan and parametrize the region in polar coordinates by $(x,y)=(r \cos t, r \sin t)$, for $r\in [0,1)$ and $t \in (0,\pi/2)$. Then $f(x,y)=r^2(\cos^2 t - \sin^2 t, 2 \sin t \cos t) = r^2(\cos 2t, \sin 2t)$. So, the image is the (open) half-disk.