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?

$i$ 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.
