Let $H=\{z=x+iy |y>0\}$ be upper half plane and $D=\{z \in \mathbb{C}||z|<1\}$ be the open unit disc. suppose that f is Mobious transformation, which maps H onto D. Suppose $f(2i)=0$. Which of the following is\are true?
- f has simple pole at $z=-2i$.
- f satisfies $f(i)\overline{f(i)}=1$.
- f has essential singularity at $z=-2i$ .
- $|f(2+2i)|=\frac{1}{\sqrt{5}}$.
How can one find Mobious transformation from H onto? what are the results behind it? I think that Once I know Mobious transformation it is easy to verify these four options for me. Please give me some hint. Thanks in advance.
$f(z)$ is a Möbius transformation : $f(z)= a\frac{z+b}{z+d}$
$f(2i) = 0$ means that $f(z) = a\frac{z-2i}{z+d}$
$f(z)$ is biholomorphic $H \to D$ means (by the open mapping theorem / maximum modulus principle) that it maps $\partial H $ to $\partial D$, i.e. $|f(x)| = 1$ for $x \in \mathbb{R} \cup \{\infty\}$,
so that $|f(\infty)|=|a|=1$ and $f(z) = e^{i \theta}\frac{z-2i}{z+d}$,
$|f(0)| = 1$ means that $|d| = 2$, and $|f(1)| = 1$ means that $d = 2i$, hence $$f(z) = e^{i \theta}\frac{z-2i}{z+2i}$$