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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!

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2 Answers

up vote 3 down vote accepted

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 $f$.

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thanks for the hint, I tried to figure out the answer. Hopefully it is correct. By the way, you mentioned "two annuli are conformally equivalent if and only if they have the same ratio of outer to inner radius", can you give a reference for this result? Thanks! –  ougao Dec 2 '12 at 4:52
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e.g. Theorem 14.22 in Rudin, "Real and Complex Analysis" –  Robert Israel Dec 2 '12 at 5:22
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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 $f: A\rightarrow B$.

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