I have the transformation $$ f(x, y) = (x^2-y^2, 2 x y) $$ and the region $A = \{(x, y): x>0 \}$. I need to find the image of $A$ under $f$,i.e., $B=f(A)$.
I think A is invariant under the transformation, but using ParametricPlot in Mathematica shows otherwise. Is there a systematic way to arrive at the correct region $B$?
You have a function from $\mathbb R^2\to\mathbb R^2$, so let's apply the substitution for polar coordinates $\begin{cases}x=\rho\cos\theta\\y=\rho\sin\theta \end{cases}$.
The set $A=\mathbb R^+\times\mathbb R$ is defined in the complex plane taking $\rho\in(0,+\infty),\theta \in(-\pi/2,\pi/2)$ and the image of the set $A$ is given by $f\Big(\{z\in\mathbb C:\rho\in(0,+\infty),\theta \in(-\pi/2,\pi/2)\}\Big)=\mathbb R^2\setminus\{(x,y)\in\mathbb R^2:x<0,y=0\}$.
Observe that (in order to have an idea of what's happening "near" the $y$ axis) $$\sqrt{\varepsilon^2+1}(\cos(3\pi/2+\alpha_{\varepsilon})+i\sin(3\pi/2+\alpha_{\varepsilon}))\overset{f}\mapsto (\varepsilon^2+1)(\cos(\pi+\bar\alpha_{\varepsilon})+i\sin(\pi+\bar\alpha_{\varepsilon}))$$