# 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.

• This is great. I thank you very much for the insight. All the best. Jan 6 '17 at 19:50

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.

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 $$w=\frac{i-z}{z+i}$$

• You can't choose the third mapping arbitrarily, because $w$ has to map the upper half-plane to the interior of the unit disk. Your choice of $0$ to $1$ works here, but it looks like blind luck! Jan 6 '17 at 19:56