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I hope that this (probably) naive question will not bother those experts. Anyway, please allow me to ask this question here:

We set $$\mathbb{A}(1/2, 2) = \Big\{z \in \mathbb{C}: 1/2 < |z| < 2\Big\}.$$

Let $p_1, \dots, p_n, z_1, \dots, z_n$ be any distinct points in $\mathbb{A}(1/2, 2)$.

Does there exist always a meromorphic function $\varphi$ on $\mathbb{A}(1/2, 2)$ such that $p_1, \dots, p_n$ are the simple poles of $\varphi$ and $z_1, \dots, z_n$ are its simple zeros, and $\varphi$ has not other zeros and poles. Moreover, $$| \varphi(re^{i \theta})| = 1, a.e. \theta \, \text{ for $r \to 1/2^{+}$ and $r \to 2^{-}$ ? } $$

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Under just slightly stronger condition, namely that $\log|\phi(re^{i\theta})|\to 0$ in $L^1$, as $r\to 2$ and $r\to 1/2$, the answer is "no".

One (real) condition must be satisfied, and this condition is $$\prod_{k=1}^n|z_k|=\prod_{k=1}^n|p_k|.$$ It follows from Jensen's formula which can be written for your ring as $$\int_0^{2\pi}\left( \log|\phi(2e^{i\theta})|-\log|\phi(e^{i\theta}/2)|\right)d\theta=2\pi\log\frac{|z_1\ldots z_n|}{|p_1\ldots p_n|}.$$ If this condition is satisfied, such function exists and can be constructed as an infinite product.

Notice that if $|\phi(re^{it})|\to 1$, as $r\to 2$ and $r\to 1/2$, your function, when exists, extends to $C\backslash\{0\}$ by reflection. This suggests how to construct the infinite product.

If my condition is not satisfied, and only $|\phi(re^{it})|\to 1$ almost everywhere, the answer is probably yes (with some very strong singularities on both circles), but are you sure you really need such a function?

EDIT. Here is the product construction. It will be more convenient to work in the ring $A=\{ z:1<|z|<4\}$. Set $h=16$. Put $$f(z)=\prod_{0}^\infty(1-h^nz)\prod_1^\infty(1-h^n/z).$$ Verify that $f(hz)=-(1/z)h(z)$ and $f(1/z)=-(1/z)f(z)$. This is a direct computation. The zeros of $f$ are at $h^n$, $-\infty<n<\infty$. Now set $$\phi(z)=\prod_{k=1}^n\frac{f(z/z_k)f(z\bar{z}_k)}{f(z/p_k)f(z\bar{p}_k)},$$ where $z_k$ and $p_k$ are your zeros and poles. Verify that this function has correct zeros and poles in the ring $A=\{ z:1<|z|<4\}$. Then verify that $\phi(1/\bar{z})=1/\overline{\phi(z)}$, which implies that $|\phi(z)|=1$ when $|z|=1$. To verify the last thing use the functional equation $f(1/z)=-(1/z)f(z)$ and the condition that $|z_1,\ldots,z_k|=|p_1,\ldots,p_k|$. The condition that $|\phi(z)|=1$ for $|z|=4$ is verified similarly.

Literature: any good course of eliptic fnctions, for example N. Akhiezer, or Whittaker Watson, chapter on theta functions. On Jensen's formula: any good course of Complex Analysis, for example Ahlfors.

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  • $\begingroup$ Dear Prof. @Alexandre Eremenko, thanks a lot. Your answer clarifies the situation a lot. Would you please tell me where I can find a reference about it (probably to you, it is obvious). I wonder whether the infinite product will be explicit and especially need to know the speed of convergence of to 1 of the modulus of that product. $\endgroup$ – Yanqi QIU Nov 9 '14 at 4:55
  • $\begingroup$ Many thanks. I learned a lot from your answer, I will take a look at one the books you mentioned. $\endgroup$ – Yanqi QIU Nov 9 '14 at 6:53

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