Let $f: \mathbb{R^2} \to \mathbb{R^2}$ be defined by the equation
$$f(x,y)=(x^2-y^2,2xy)$$
Let $A$ be the set consisting of all $(x,y)$ with $x \gt 0$.
and $g: \mathbb{R^2} \to \mathbb{R^2}$ by the equation
$$g(x,y)=(e^x\operatorname{cos}y,e^x\operatorname{sin}y)$$
Let $B$ be the set consisting of all $(x,y)$ with $0\lt y\lt 2\pi$.
What are the sets $C=f(A)$ and $D=g(B)$?
$D$ is easier to think because I can think of $e^x$ as the radius and then there would be circles without the line $x\ge 0$. However, I'm not sure how to formally prove this idea. Meanwhile, for $C$, I'm having trouble interpreting the image of the set $A$. I could think of the complex function $z^2$ but I don't know how to find the image of the set with only positive real part. Less so in the $\mathbb{R^2}$, I don't know how to show the image. I would appreciate any help.