Finding a Mobius transformation that maps the upper half-plane to the inside of a unit-disk, such that point $i$ is mapped to $0$ and $\infty$ to $-1$ Finding a Mobius transformation that maps the upper half-plane $\{Im(z)>0\};z \in \mathbb{C}$ to the inside of a unit-disk, such  that point $i$ is mapped to $0$ and $\infty$ to $-1$.
Okay, to be clear, I know that a Mobius transformation $w$ is of the form:
$$w=\frac{az+b}{cz+d};ad-bc\neq0.$$
I am very aware of what a unit disk is. I have done assignments like finding an mobius transformation that maps some points ($\mathbb{C}$) $a,b,c$ to $d,e,f$. -Where these points were not infinity.
But nothing like this problem that I have here? How is this done? I think I just need one more point and know it's picture to be able to figure is out. But how?
 A: $i$ is mapped to $0$, i.e. $\dfrac{ai+b}{ci+d} = 0$. So $b=-ai$.
$\infty$ is mapped to $-1$, i.e. $\dfrac{a}{c}=-1$. So $c=-a$.
We can choose $a$ to be equal to $1$. So $b=-i$ and $c=-1$, giving for some $d$
$$w(z)=\dfrac{z-i}{-z+d}$$
We want to choose $d$ so that $w$ maps the upper half-plane to the interior of the unit disk. $|z-i|$ is the distance from $z$ to $i$; and the upper half-plane is the set of points that are closer to $i$ than to $-i$. So if we choose $d=-i$, we will have $0 \le |w(z)| < 1$ for $z$ in the upper half-plane, as required.
A: One very useful Mobius transformation to remember is the Cayley transform defined by
$$ f(z)=\frac{z-i}{z+i}$$
This maps the upper half plane to the open unit disk because the upper half plane is precisely the set of points in $\mathbb{C}$ which are closer to $i$ than to $-i$. Note also that $f(i)=0$.
This map doesn't quite fit your requirements because $f(\infty)=1$, but $-f(z)$ works instead.
A: When wanna map three points $\{z_1,z_2,z_3\}$ to $\{w_1,w_2,w_3\}$, use the formula 
$$\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}=\frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}$$
In this practice we have two points and need an extra point, so we choose $0$ to $1$, then we map 
$\{i,\infty,0\}$ to $\{0,-1,1\}$. If one of the points were the point at $\infty$ (here $z_2=\infty$) we delete it from formula:
$$\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}=\frac{z-z_1}{z-z_3}$$
$$\frac{w-0}{w-1}\times\frac{-1-1}{-1-0}=\frac{z-i}{z-0}$$
after siplifying
$$w=\frac{i-z}{z+i}$$
