Non-existence of a bijective analytic function between annulus and punctured disk Suppose $A=\{z\in \mathbb{C}: 0<|z|<1\}$ and $B=\{z\in \mathbb{C}: 2<|z|<3\}$. 
Show that there is no one -to-one analytic function from A to B. Any hints? Thanks!
 A: In fact, two annuli are conformally equivalent if and only if they have the same ratio of outer to inner radius.    But this case is easier than the general one.  Hint: if $f: A \to B$ is such a function, $0$ is a removable singularity of $f$.
A: Based on Robert's hint above. I tried to figure out the following answer.
First, since $f$ is well-defined (i.e., the value $f(z)$ is bounded when $z$ is around zero point), then 0 must be a removable singularity of $f$. 
Then, consider $w=f(0)$.
Case 1: $w\in B$. It is impossible, since then, pick $0\neq z\in A$, such that $f(z)=w$ (since $f$ is onto). Then, we can find two disjoint neighborhood $A_0, A_z\subset \bar{A}$ of $0$ and $z$ respectively such that $f(A_0)\cap f(A_z)$ is a neighborhood of $w$ in $B$. That is a contradiction, since for any $w'\neq w\in f(A_0)\cap f(A_z)$, we have find two distinct points $z_1\in A_0, z_2\in A_1$ such that $f(z_1)=f(z_2)=w'$, which contradicts the assumption of the injectivity of $f$.
Case 2, $w\in \partial B$. This is also impossible. Since this contradicts the open mapping theorem, stated in page 204 in Rudin's "Real and Complex Analysis", which says that if we pick a small neighborhood $A_0\subset \bar{A}$, then $f(A)$ is either an open set of $\mathbb{C}$ or a point, which is impossible in our case.
So, we can not find an analytic bijective map $f: A\rightarrow B$.  
