Sketching the image of a given function Given the upper half-plane $\cal H$ and the map $$\varphi:\cal H\rightarrow\mathbb{C},~z\mapsto\frac{z-i}{z+i}.$$
Can somebody sketch the image $\varphi(S)$ with $$S=\{z\in{\cal H}:|\text{Re}(z)|<1/2,~ |z|>1\}?$$
I know that the image is included in the unit disk. And I guess that the boundary is some kind of triangle. Does this 'triangle' touches the boundary of the unit disk? Where is the point $i\infty$ maped to? 
 A: This being a Mobius transformation, $\varphi$ maps straight lines or circles to straight lines or circles.  A straight line or circle is determined by $3$ points, so take three points on it, find their images, and see what is the straight line or circle through those points.  
If $z$ is real, $z - i = \overline{z+i}$, so $\varphi(z) = 1$.  Thus $\varphi$ maps the real line to the unit circle.  So it's not true that the image is in the unit circle.  The upper half plane maps into the unit circle, the lower half plane outside.  
A: Let $$w=u+iv=\frac{z-i}{z+i}$$
Rearranging, we have $$z=i\left(\frac{1+w}{1-w}\right)$$
The region $S$ is defined by three conditions. 
Firstly, $|z|>1\implies|1+w|>|1-w|\implies u>0$
Now when $z$ is expressed in terms of $u$ and $v$ and simplified, we get $$z=\frac{i(1-u^2-v^2)-2v}{(1-u)^2+v^2}$$
So applying the second condition, that $Im(z)>0$ leads to the inequality $$u^2+v^2<1$$
Finally we also have $Re(z)<\frac 12$ and this leads to the inequality $$(u-1)^2+(v+2)^2>4$$
So the required region lies to the right of the imaginary axis, inside the unit circle and outside the circle whose centre is $(1,-2)$ with radius $2$.
A: Here is a sketch of the unit circle containing region $\varphi(S)$ which is the interior of the "triangle";


*

*The vertical line segment $[i(-2+\sqrt{3}),i(2-\sqrt{3})]$ is the image of the arc of circle. 

*The circle arcs are the images of vertical boundaries. Due to the fact that $\varphi(z)$ can be written under the form $\dfrac{1-\frac{i}{z}}{1+\frac{i}{z}}$, $\varphi(i\infty)$ is clearly $1$.
Remark 1: $S$ is known to be a fundamental region for the action of a $PSL(2,\mathbb{Z})$ on $\mathbb{C}$ on $H$ (the upper plane Poincaré space) and is pictured here 
Remark 2: for a proof that $\varphi(S)$ is like that, see the answer by @David Quinn.

