This was one of the problems on a previous year's Complex Analysis final exam.

Assume $f\in \mathcal O (\mathbb H )$, non-constant, and $f(\frac {i}{\sqrt n})=0, \forall n\in \mathbb N$. Prove that $f$ takes unbounded values.

What I tried so far: I tried to argue that the point $z=0$ had to be an essential singularity since the function cannot be continued to be holomorphic there (taking the value $0$), for then it would be forced to be identically the $0$ function (which it is assumed not to be). Then i squared the input domain to argue that the essential singularity must take on unbounded values in the upper-half plane somewhere near $z=0$. But, I think my reasoning is wrong because this may not be an isolated singularity at all and may be a point in the branch cut of a holomorphic function or something. I'm wondering if someone can write me up a nice proof and/or explanation about how to tackle this problem, thanks.


1 Answer 1


Using a fractional linear transformation, change the domain to the unit disk. Make it so that the center of the disk is not a zero of the new function (call it $g$). Let $\{a_n\}$ be the zeros of $g$. Since $g$ is bounded, Jensen's formula implies that for every $r\in (0,1)$ $$ \sum_{|a_k|< r} \log \frac{r}{|a_k|} = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta - \log |g(0)| $$ where the right hand side is bounded independently of $r$. Letting $r\to 1$, conclude that $$ \sum_{k} \log \frac{1}{|a_k|}<\infty $$ or equivalently $$ \sum_{k} (1-|a_k|)<\infty $$

But the sequence $i/\sqrt{n}$ is transformed into a sequence $\{b_n\}$ such that $1-|b_n|$ is comparable to $1/\sqrt{n}$.

  • $\begingroup$ I'm not sure I understand the whole process. Could you spell it out a little more? Also, I realized after the fact that we didn't go over Jensen's formula yet in the class this quarter so I'm kind of unfamiliar with it. $\endgroup$ Mar 18, 2015 at 18:40

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