Mobius transformations from intersection of circles to two straight lines My textbook is pretty useless on Mobius transformations, and I just wanted to check this reasoning was correct!
I want to find a Mobius mapping from the region $$\{z:|z-1|\lt\sqrt{2}, |z+1|\lt \sqrt{2}\}$$ to the region $$\{z:\frac{3\pi}{4}\lt argz \lt \frac{5\pi}{4}\}$$I know that the two circles in the first region intersect at $i$ and $-i$ so I consider a Mobius transformation $z \mapsto \frac{z+i}{z-i}$ which takes $-i$ to $0$ and $i$ to $\infty$. I then tested two points, one on the left circline that bounds the two circles' intersections (namely $1-\sqrt{2}$) and one on the right circline that bounds the two circle's intersections (namely $-1+\sqrt{2}$). From here I see that $1-\sqrt{2}$ goes to $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$,so onto the half line $y=x$ for $y,x\lt 0$. And similarly, $-1+\sqrt{2}$ gets mapped to the line $x=-y$ for $x \lt 0$ and $y\gt 0$. These two boundaries seem to bound precisely the region we want. 
Thus I think this mapping, $z \mapsto \frac{z+i}{z-i}$ is a Mobius mapping from the first region to the second. Is this correct? 
 A: If I'm not mistaken, you've just shown that the boundary of one region goes to the boundary of the other, and you still need to show that the interior goes to the interior.  The argument you've made so far leaves open the possibility that the region maps to $${5\pi\over4}<\arg z< {11\pi\over4}$$
Just pick a point says $0$ in the original region.  It maps to $-1$, which is inside the region you want.  Now the argument is complete.
A: Looking at the original picture, the 2 circles split your region into 4 parts, the intersection of the two circles, one circle - intersection, other circle -intersection, and the complement of the union of the two circles.
These 4 regions will be mapped to the 4 sectors formed by the two lines $y=x$ and $y=-x$, therefore, the intersection will be one of these 4 sectors. 
By testing the point $z=0$, we observe, that $0$ which lies in the intersection gets mapped to $\frac{i}{-i}=-1,$ which is in the required sector. 
In general, if $\alpha$ and $\beta$ are any two points in $\mathbb{C}$, then $w=\frac{z-\alpha}{z-\beta}$ maps any circle containing $\alpha$ and $\beta$ to straight lines through the origin in the $w$-plane. 
