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Let $f$ be analytic and bounded on $\{x+iy\in\mathbb{C}:|y|<\frac{\pi}{2}\}$. Suppose $f(\ln n)=0$ for all $n\in\mathbb{N}$. Show that $f$ is identically 0.

I tried to perform some transformations to end up in the unit disk to see if I could get anything. First I translated by $i\frac{\pi}{2}$ to get to the strip $\{x+iy\in\mathbb{C}:0<y<\pi\}$. Then I exponentiated to get to the upper half-plane. Finally I used $z\mapsto\frac{i-z}{i+z}$ to get to the unit disk. Under the composition of these maps, $\ln n$ in the original region corresponds to $\frac{1-n}{1+n}$ in the unit disk. I can't proceed from here.

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Let $\phi(z) = \ln \frac{1+z}{1-z}$. $\phi$ is a conformal map of the open unit disk onto the set in question. Hence $\tilde{f} = f \circ \phi$ is a bounded, analytic function on the open unit disk.

A quick calculation shows that $\phi(\frac{n-1}{n+1}) = \ln n$, hence $\alpha_n =\frac{n-1}{n+1}$ is a zero of $\tilde{f}$ for every $n$. Furthermore, $\sum_n (1-|\alpha_n|) = \sum_n \frac{2}{1+n} = \infty$.

Theorem 15.23 in Rudin's "Real & Complex Analysis" asserts that if $f$ is analytic and bounded on the unit disk, not identically zero, and $\beta_n$ are the zeroes of $f$ listed according to their multiplicities, then $\sum_n (1-|\beta_n|) < \infty$.

It follows that $\tilde{f}$ must be identically zero, and since $\phi$ is onto, that $f$ is identically zero.

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The strip $G$ is simply connected, then by Riemann Mapping Theorem, there is 1-1 analytic function $g$ maps the strip onto the unit disk $\mathbb D$.

Define $h(z)=f\circ g^{-1}(z)$, clearly $h$ is analytic on $\mathbb D$ and has infinitely many zeros in $\mathbb D$. Then by Identity Theorem, $h\equiv 0$. Since $g$ is not identically zero, $f\equiv 0$.

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There are non-constant (even bounded) holomorphic functions on the unit disk that have infinitely many zeros. This argument is not sufficient. –  Daniel Fischer May 19 at 20:00
    
@DanielFischer, yes, you are right. Infinitely many zeros don't imply existence of a limit. –  Falang May 19 at 20:22

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