Mobius transformation that maps the upper half plane conformally onto the open unit disc.

Question

Let $H = \{z=x+iy \in \Bbb C:y>0 \}$ be the upper half plane and $D=\{z \in \Bbb C:|z|<1 \}$ be the open unit disc. Suppose that $f$ is a Mobius transformation, which maps $H$ conformally onto $D$. Suppose that $f(2i)=0$. Pick each correct statement from below.

$1.$ $\ f$ has a simple pole at $z=-2i$.

$2.$ $\ f$ satisfies $\ f(i) \overline {f(-i)} = 1$.

$3.$ $\ f$ has an essential singularity at $z=-2i$.

$4.$ $\ |f(2+2i)|=\frac {1} {\sqrt 5}$.

I have tried to proceed in the following way $:$

Since $f$ is a Mobius transformation so there exist complex constants $a,b,c,d$ with $ad-bc \neq 0$ such that $f(z)=\frac {az+b} {cz+d}$, $z \in H$. Now since $f(2i)=0$ so we have $b=-2ai$. So $f(z)=\frac {a(z-2i)} {cz+d}$,$z \in H$.Now I got stuck. How should I proceed from here? Please help me in this regard.

Thank you very much.

Answer 1

An example of such a Mobius transformation is given by $f(z)=\frac{z-2i}{z+2i}$. If $g$ is another such Mobius transformation, then $g\circ f^{-1}$ is an automorphism of the open unit disk and $g\bigl(f(0)\bigr)=0$. Therefore, $g\circ f^{-1}=\omega\operatorname{id}$ for some $\omega$ such that $|\omega|=1$. It follows that$$(\forall z\in\mathbb{H}):g(z)=\omega f(z)=\omega\frac{z-2i}{z+2i}.$$So, the answers are:

1. True.
2. True.
3. False.
4. True.