I am dealing with the region $y-x>0$. I would like to transform this region by rotating it to end up with the upper half plane.
So far I have tried to create the new region by taking $\tilde{y}=y-x$. Then the transformed region would be $\tilde{y} >0$ which is what I want. However, I'm confused on how to deal with the $\tilde{x}$ in this new transformed space. All I know so far about this coordinate is that $\tilde{x}=y-x=0$ and that it is essentially just the original $x$ axis rotated by 45 degrees. I am looking for help explicitly defining $\tilde{x}$. If there is something obvious I am missing please let me know. Thanks!
To make computations simpler the point $(x,y)$ in the plane before transforming can be sent to the point $(u,v)=(y+x,y-x).$ This is in fact the result of rotating by $-\pi/4$ but then stretching by a factor of $\sqrt{2}.$
Then if originally $y-x>0$ then the altered second coordinate $v$ is greater than $0$ as desired. The fact that the new first coordinate is $y+x$ might make one think it must be positive, but if one finds the inverse of our map it is $(x,y)=((u-v)/2,(u+v)/2.)$ So for example the point $(x,y)=(-2,-1)$ maps to the point $(u,v)=(-3,1)$ showing a case where the new first coordinate is negative.
To explain how I got my $(u,v)$ : First I used complex numbers to do a rotation through $-\pi/4$ using the fact that angles add when complex numbers multiply, so that multiplying $x+i y$ by $e^{-i \pi/4}$ gives the complex number $(u,v)=(y+x)/\sqrt{2}+i(y-x)/\sqrt{2}.$ [to do this i wrote $e^{i (-\pi/4)}$ in the form $1/\sqrt{2}-i(1/\sqrt{2}).$]
Then thinking the squareroot of two made it complicated I just dropped it, hence the stretch by that factor.