# Holomorphic function $F:\Bbb H\to \Bbb C$ having the bounded sequence $\{ir_n\}$ as zeros.

Stein and Shakarchi, Complex Analysis, Chapter 8 Problem 5.

1. Suppose that $$F:\mathbb{H}\to\mathbb{C}$$ is holomorphic and bounded. Also, suppose $$F(z)$$ vanishes when $$z=ir_n$$, $$n=1,2,3,\ldots,$$ where $$\{r_n\}$$ is a bounded sequence of positive numbers. Prove that if $$\sum r_n=\infty$$ then $$F=0$$.
2. If $$\sum r_n<\infty$$, it is possible to construct a bounded function on the upper half-plane with zeros precisely at the points $$ir_n$$.

There is something weird about the first part. If the sequence $$\{ir_n\}$$ is infinite then it has a convergent subsequence since it's bounded. Hence the zeros of $$F$$ accumulate in $$\Bbb H$$ and $$F$$ is zero. Am I missing something here?

If $$\{ir_n\}$$ accumulates at a point other than $$0$$, then $$F=0$$ is trivial as you said. But the problem is requiring you to show that if $$\{ir_n\}$$ accumulates at $$0$$ and the convergence $$r_n\to 0$$ is "slow" enough to make $$\sum_n r_n =\infty$$, then $$F$$ must be $$0$$. Put differently, if $$F\neq 0$$ is bounded on the upper half plane and $$\{z_n\}$$ are zeros of $$F$$, then it must be that $$\Im(z_n)\to 0$$ fast enough to make $$\sum_n \Im (z_n)<\infty.$$
• That's the trick of this problem. How can we transform the domain into the unit disk without losing analyticity of $F$? (Hint)Maybe you can use $z = \frac{aw+b}{cw+d}$ or something like this. – Myeonghyeon Song Dec 16 '18 at 20:38
• Use the map $T:\Bbb D\to \Bbb H$ defined by $z\mapsto i\frac{1-z}{1+z}$? – UserA Dec 16 '18 at 20:40
The sequence $$(ir_n)$$ accumulates at a point of $$\mathbb{C}$$, but not necessarily at a point of $$\mathbb{H}$$. Indeed, if $$r_n\to 0$$ then they accumulate only at $$0$$, which is not in $$\mathbb{H}$$.