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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.

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  • $\begingroup$ @KaviRamaMurthy why $(1)$ is false? Consider the Mobius transformation $z \mapsto \frac {z-2i} {z+2i}$. It is map that sends $H$ onto $D$ but clearly $-2i$ is a pole of $f$. $\endgroup$ – D_C Jun 12 '18 at 8:41
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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.
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  • $\begingroup$ $(2)$ is true according your function. $\endgroup$ – D_C Jun 12 '18 at 8:44
  • $\begingroup$ @D_C I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jun 12 '18 at 8:45
  • $\begingroup$ But I can't understand the reasoning you have used here. Why $g \circ f^{-1}$ is an automorphism of the open unit disc? As a result of that why does it follow that $g \circ f^{-1}=\omega\ \mathrm {id}$? $\endgroup$ – D_C Jun 12 '18 at 8:46
  • $\begingroup$ @D_C Mobius transformations are injective. So, $f$ and $g$ are holomorphic bijections from $\mathbb H$ onto $\mathbb D$ and therefore $g\circ f^{-1}$ is a holomorphic bijection from $\mathbb D$ onto itself. Besides, it is a Mobius transformation. The Mobius transformations from $\mathbb D$ onto itself are those of the form $z\mapsto\omega\frac{z-a}{\overline az-1}$, with $|\omega|=1$; there's a proof here. In this case, it is a transformation zuch that $0$ is mapped into itself. So, $a=0$, and we get $(g\circ f^{-1})(z)=\omega z$. $\endgroup$ – José Carlos Santos Jun 12 '18 at 9:05
  • $\begingroup$ Many many thanks for your help @JoséCarlosSantos. The pdf in the link is very helpful. $\endgroup$ – D_C Jun 12 '18 at 9:55

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