I'm self-studying some complex analysis, and apparently holomorphic bijections between two annuli exist precisely when the ratios of the radii are the same. More exactly, if $A_{\sigma,\rho}=\{z\in\mathbb{C}:\sigma<|z|<\rho\}$, then there is a holomorphic bijection between $A_{\sigma,\rho}$ and $A_{\sigma',\rho'}$ iff $\rho/\sigma=\rho'/\sigma'$.

Is there a reference where this fact is proven? Or can a proof be included here if it's not overly involved? Thanks.

  • $\begingroup$ This is exercise 1 on page 249 of Ahlfohrs, after both Riemann mapping and a fairly deep look at harmonic functions and the Dirichlet problem. The section is titled Canonical Mappings of Multiply Connected Regions. $\endgroup$ – Will Jagy Apr 18 '12 at 19:14
  • $\begingroup$ You should find this in your favorite complex analysis book. Mine is Greene & Krantz: Function Theory of One Complex Variable (It's in chapter 7) $\endgroup$ – Brett Frankel Apr 18 '12 at 19:22

This result (sometimes called Schottky's theorem) can be proved without any heavy machinery like Riemann mapping or even Schwarz reflection. I give two versions of the proof, with comments at the end.

Claim. If there exists a bijective holomorphic map $f: A_{r,R}\to A_{s,S}$, then $S/s\ge R/r$.

Note that we actually get equality by considering the inverse.

Proof. Normalize to $r=s=1$. By composing $f$ with inversion, we can make sure that $|f(z)|\to 1$ as $|z|\to1$. For $1<t<R$ let $A(t)$ denote the area within the Jordan curve $f(\{z:|z|=t\})$. There is a standard way to relate the area to the coefficient of the Laurent series $f(z)=\sum_{n\in\mathbb Z} c_n z^n$: namely, use Green's formula for area in complex notation. $$A(t)=\frac{1}{2i} \int_{0}^{2\pi} f(te^{i\theta}) \bar f(te^{i\theta})\, d\theta = \pi \sum_{n\in\mathbb Z} n|c_n|^2t^{2n}$$ Since $A(1+)=\pi$, we have $$(1)\qquad\qquad \sum_{n\in\mathbb Z} n|c_n|^2=1.$$ From here the proof can proceed in at least two ways.

Version I: stick to area. Using (1), write $$A(t)-\pi t^2=\pi t^2 \sum_{n\in\mathbb Z} n|c_n|^2(t^{2n-2}-1)\ge0$$ where the inequality holds term-wise. Hence $A(R-)\ge \pi R^2$, which implies $S\ge R$. QED

Version II: use $L^2$ norm. For $1<t<R$ define $$U(t)=\frac{1}{2\pi t} \int_{|z|=t} |f(z)|^2|dz|$$ Again, this can be put in terms of coefficients either by direct computation with $|f|^2=f\bar f$, or by Parseval's identity: $U(t)= \sum_{n\in\mathbb Z} |c_n|^2 t^{2n}$. Since $U(1+)=1$ and $$U'(t)-2t= 2t\sum_{n\in\mathbb Z} n|c_n|^2 (t^{2n-2}-1)\ge 0 $$ it follows that $U(R-)\ge R^2$. But $|f|\le S$, which implies $S\ge R$. QED

Comments: the idea of relating coefficients to area as in (1) goes back to Gronwall. In the context of this problem it was used by Pólya and Szegő in Problems and Theorems in Analysis II, part IV, problem 83 on page 15. Their solution is essentially Version I. I learned it from a short note by Pietro Poggi-Corradini.

The idea of using $L^2$ norm instead of area may look counterproductive, since Version II is longer. But it is a more robust argument, which works for harmonic maps as well. Robert Burckel and Pietro pointed out Version II to us as a simplification of $L^2$ argument specialized to holomorphic maps.

  • $\begingroup$ This is awesome. Did this argument appear in one of their papers? $\endgroup$ – t.b. Jul 28 '12 at 7:40
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    $\begingroup$ @t.b. I rewrote the proof[s] and attempted to put their history in better perspective. $\endgroup$ – user31373 Jul 28 '12 at 17:13
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    $\begingroup$ Thanks again for this great answer! $\endgroup$ – t.b. Jul 28 '12 at 18:10
  • $\begingroup$ Hey thanks LVK! $\endgroup$ – Dedede Aug 28 '12 at 20:39

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