How to visualize bilinear transform of the form $S(z) = \frac {T}{2} \frac {z+1}{z-1}$ Note that this is the bilinear transform from a z-domain as appears in Z-transform to s-domain in Laplace transform
Recall that bilinear transform has form  $M(z) = \frac{az+b}{cz+d}$ with and has to satisfy conditions $M(\infty) = \frac{a}{c}$, $M(0) = \frac{b}{d}$, $M(-d/c) = \infty$
In this case, we have $S(z) = \frac {T}{2}  \frac {z+1}{z-1}$ where  a = T, b = T, c = 2, d = -2
My question is how should I interpret the three conditions on bilinear transform namely 
$S(\infty) = \frac{T}{2}$, $S(0) = \frac{T}{-2}$, $S(-(-2)/2) = \infty$
This is currently how I visualize the conditions:
So z = $\infty$ is mapped to T/2 on the real line of the laplace domain
z = 0 is mapped to -T/2 on the real line
z = 1 is mapped to infinity

But this drawing (straight from my interpretation of these conditions) is not making any sense to me! I am not sure where z = 0, and 1 located on the Z-domain.
Can someone explain how I should understand the three conditions of this bilinear transform? How should I redraw this picture so the graph is consistent with the conditions?

If $T=\sqrt2$,then your map is just a rotation of the Riemann sphere (a bilinear map is a rotation of the Riemann sphere when the coefficient matrix is unitary: http://www.math.msu.edu/~shapiro/Pubvit/Downloads/RS_Rotation/rotation.pdf)
So think of the Riemann sphere, rotating it so infinity goes to $1/ \sqrt 2$ and 0 goes to $-1/\sqrt 2$ (and 1 goes to $\infty$). Then think about the complex plane and stretch it by a factor of $T/\sqrt 2$. 
