How to find range of a function $f : \mathbb{R}^2\to \mathbb{R}^2 $ 
How to find range of a function  $f : \mathbb{R}^2\to \mathbb{R}^2 $
defined by
$f(x,y)=(x^2-y^2, 2xy)$

I am new to calculus of several variables and I have no idea on how to solve such questions. I tried the following way:
We set $f(x,y) = (x^2-y^2, 2xy) = (h,k)$ but it is getting very dirty solving for $x$ and $y$ from this. However, is this the correct method?
 A: Yes, your approach is correct. In this case, I would use $$x=r\cos\theta\\y=r\sin\theta$$
Then $$f(r,\theta)=r^2(\cos2\theta,\sin2\theta)$$
It's easy to see that $$r^4=h^2+k^2$$ and $$\tan2\theta=\frac kh$$ These have solutions for any $(h,k)\in\mathbb R^2$, so the range is $\mathbb R^2$.
A: This is not so much a calculus question as a trick question (maybe not a trick but not really part of the usual theory) in my opinion.
We want to find all points of the form $(x^2-y^2,2xy)$.
We do some variable bamboozlement, the right coordinate seems easier, so lets say $2xy =a$ when $a$ is not $0$, then we get $y = a/2x$.
now we want to find the possible values of $x^2-(a/2x)^2$, it turns out it is all values. To see this notice the function is continuous on $(0,\infty)$, the limits when $x$ goes to $0$ is $-\infty$ and the limit when $x$ goest to $\infty$ is $\infty$.
for when $a$ is $0$ you can also cover all values, when you want $x^2-y^2$ to be positive you make $y=0$, when you want $x^2-y^2$ to be negative you make $x=0$ and when you want $x^2-y^2$ to be $0$ you make both equal $0$.
We conclude the range is all of $\mathbb R^2$.
A: This function applied to the coordinates of the complex number $x+iy$ gives us the coordinates of $(x+iy)^2$. This function from $\mathbb C$ to $\mathbb C$ is surjective as a consequence of the fundamental theorem of algebra or by noticing the function squares the modulus and doubles the argument (which is clearly surjective).
