Given $A(a,b) = \{ z \in \mathbb{C} : a < \left| z \right| < b \} $, with $0<a<b$ then $$ A(a,b) \simeq A(a',b') $$ if and only if $a/b= a'/b'$.
I don't get some parts of the following proof:
Let $ \mathfrak{h} = \{ z \in \mathbb{C} : \Im (z) > 0 \} $. Consider the action of $ \mathbb{Z} $ determined by the action of $1$ and suppose that $1$ act by multiplication of $\lambda > 1$. Then we have that $z \mapsto \exp \left( 2 \pi i \frac{ \operatorname{Log}(z)}{\log \lambda} \right) $ induces a biholomorphism $ \mathfrak{h} / \mathbb{Z} \simeq A\left( \exp \left( - \frac{2 \pi^2}{\log \lambda} \right),1 \right)$. Then suppose that $g: A(a/b,1) \simeq A(a'/b',1) $ is a conformal map, where $ \lambda , \lambda' > 1$ are such that $ \frac{a}{b} = \exp \left( - \frac{2 \pi^2}{\log \lambda} \right)$ and $\frac{a'}{b'} = \exp \left( - \frac{2 \pi^2}{\log \lambda'} \right)$. Since $ \mathfrak{h} $ is connected then $g$ lifts to a biholomorphism $f: \mathfrak{h} \to \mathfrak{h} $. Since the diagram is commutative we deduce that $f(\lambda z) = (\lambda')^n z $ for some $n \in \mathbb{Z}$, and since $f$ is an automorphism of $\mathfrak{h} $ is of the form $z \mapsto \frac{az+b}{cz+d}$ for some $a,b,c,d \in \mathbb{R}$ and $ ad-bc=1$, then we deduce that $f(z) =z $ and $\lambda= \lambda'$, i.e. $ \frac{a}{b} = \frac{a'}{b'}$.
My questions:
Why $z \mapsto \exp \left( 2 \pi i \frac{ \operatorname{Log}(z)}{\log \lambda} \right) $ induces a bioholomorphism ? Okay we have that if $\varphi : \mathfrak{h} \to A\left( \exp \left( - \frac{2 \pi^2}{\log \lambda} \right),1 \right)$ defined by $z \mapsto \exp \left( 2 \pi i \frac{ \operatorname{Log}(z)}{\log \lambda} \right)$ we have that this is holomorphic and the image is contained in $A\left( \exp \left( - \frac{2 \pi^2}{\log \lambda} \right),1 \right)$ and I think also that is surjective. But if we consider the quotient map $ \pi : \mathfrak{h} \to \mathfrak{h}/\mathbb{Z}$, we need to check that $\varphi $ passes to the quotient, so that if $\varphi(z) = \varphi(w) $ then we must have that $ \pi(z) = \pi(w) $ or equivalently $$\log \left| z \right| + i \arg(z) = \operatorname{Log}(z) = \operatorname{Log}(\lambda z) =\log \left| \lambda z \right| + i \arg( \lambda z) $$ but this seems to me wrong.
I don't get why $f(\lambda z) = (\lambda')^n z$.
I don't get how from $z \mapsto \frac{az+b}{cz+d}$ we deduce that $f(z)=z$, actually I'm only able to say that $ c=b=0$ and $a \lambda = d (\lambda')^n$.