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, $$| \varphi(re^{i \theta})| = 1, a.e. \theta \, \text{ for $r \to 1/2^{+}$ and $r \to 2^{-}$ ? } $$


1 Answer 1


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.

  • $\begingroup$ 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, 2014 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, 2014 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.