Describe the image of the set $\{z=x+iy:x>0,y>0\}$ under the mapping $w=\frac{z-i}{z+i}$ Describe the image of the set $\{z=x+iy:x>0,y>0\}$ under the mapping $w=\frac{z-i}{z+i}$
So from this mapping , I can see that $a=1, b=-i, c=1, d=i$ thus $ad-bc=i+i=2i \not =0$ so this is a Mobius transformation. Solving for $z$ I got
$$z=\frac{i+iw}{1-w}$$
for $w=u+iv$, we have 
$$z=\frac{-2v+i(1-u^2-v^2)}{(1-u^2)+v^2}$$
so $x=\frac{-2v}{(1-u^2)+v^2}$ and $y=\frac{1-u^2-v^2}{(1-u^2)+v^2}$
Since $x>0$, $v<0$
and since $y>0$, $1-u^2 -v^2 >0$, thus $u^2 +v^2 <1$, this implies that the image is the interior of a unit circle center at the origin, but since $v<0$, we only take the negative part. 
 A: The boundaries of the domain are $y=0,x>0$ and $x=0,y>0$.  The image of $y=0,x>0$ is the lower half of the unit-circle because $\left|\frac{x-i}{x+i}\right|=1$.  Try to work out the image of the upper imaginary axis.
A: Here is my preferred method.  A linear fractional transformation maps circles to circles (on the Riemann sphere, so lines are a special case of circles).
The boundary if the set $\{z=x+iy:x>0,y>0\}$ is two arcs of circles (which are parts of straight lines in this case), namely (1) the positve real axis and (2) the positive imaginary axis.  
For (1), the circle begins at $0$, goes through $1$, and ends at $\infty$.  Our region is on the left when (1) is traced in that direction.  The images of these three points are $-1$, $-i$, $1$.  So the image arc begins at $-1$, goes through $-i$ and ends at $1$.  The unique arc of a circle that does that is lower half of the unit circle.  The region on its left will be inside the unit circle.
For (2), the circle begins at $\infty$, goes throught $i$, and ends at $0$.  When traced in that direction, our region is on the left.  The images of these three poins are $1$, $0$, $-1$.  The circle through these is the real axis.  The region on the left is below the real axis.  
Our final answer is a "lune", in this case a half-disk;  bounded below by the lower half of the unit circle and above by the segment from $-1$ to $1$.
