# Problem of the Week #44 - April 1st, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_1,z_2,\ldots,z_n,\ldots$ are its zeros (with $|z_k|<1$), then
$\sum_n(1-|z_n|)<\infty.$

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Hint:
Use Jensen's formula.

#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. Here's my solution below.

Proof: In dividing $f$ by $z^n$, if necessary, we can assume that $f(0)\neq 0$. Now, suppose otherwise in the given statement. Since $-\log x \geq 1-x$ for $x\in (0,1)$, this implies that $-\sum\log|z_n|\rightarrow \infty$ and thus $-\log(\prod_n|z_n|)\rightarrow\infty$ which now implies that $\prod_{k=1}^n|z_k|\rightarrow 0$. By Jensen's formula, for $r<1$, and $z_1,\ldots,z_n$ the zeros in $|z|<r$,
$\log|f(0)|=\sum_{k=1}^n\log(|z_k|/r)+\frac{1}{2\pi} \int\log|f(re^{i\theta})|\,d\theta.$
Since $|f|$ is bounded on the unit disc, by letting $r\rightarrow 1$ and exponentiating this formula, there is a constant $M$ so that $|f(0)|\leq e^M|\prod_{k=1}^n z_k|$, which contradicts $f(0)\neq 0$ and $\prod_{k=1}^n|z_k|\rightarrow 0$.$\hspace{1in}\blacksquare$

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