Biholomorphism between 2 non simply connected domains

Question

Is the annulus of centre $0$ and radii $1$ and $2$ biholomorphic to the punctured disc $$\{ z \in \mathbb{C} \ | \ 0<|z| <1\}$$ ? Why ?

I know the Riemann mapping theorem but here we have non simply connected domains.

And also of this one. – you can call me Al Jan 30 '14 at 12:37 — you can call me Al, Jan 30 '14 at 12:37

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

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The answer is no. The more general result is:

For $0\le r<R<\infty$ and $0\le r^\prime<R^\prime<\infty$, the annuli $$\{ z\in\mathbb{C}\,:\,r<z<R \}$$ and $$\{z\in\mathbb{C}\,:\,r^\prime<z<R^\prime\}$$ are biholomorphically equivalent if and only if $$R r^\prime=r R^\prime$$

You will find the proof probably in any beginner's text book on complex analysis.

You can read up the proof of the general statement for example here.

Your special case is easier since $r=0$. See this link given in the comment above

Note: In $\mathbb{C}$, biholomorphic equivalence is the same as conformal equivalence.

edited Apr 13 '17 at 12:20

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answered Jan 30 '14 at 12:31

J.R.

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@WLOG: It is often given as an exercise, possibly with a few hints in a complex analysis course. At least that's where I know it from. – J.R. Jan 30 '14 at 12:34
but in your lemma $r \neq 0 $, in my problem $r =0$ – WLOG Jan 30 '14 at 12:36
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@WLOG: No, I wrote $r\ge 0$. – J.R. Jan 30 '14 at 12:37

Biholomorphism between 2 non simply connected domains

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