Can You Prove the Convergence of Zeros of a Holomorphic Function?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on proving that for a holomorphic function \( f \) defined in the unit disc, which is bounded and not identically zero, the series of the distances from the zeros \( z_n \) to the boundary of the unit disc converges. Specifically, it establishes that if \( |z_k| < 1 \) for the zeros \( z_1, z_2, \ldots, z_n, \ldots \), then the sum \( \sum_n(1 - |z_n|) < \infty \). This result is significant in complex analysis and has implications for the distribution of zeros of holomorphic functions.

PREREQUISITES
  • Understanding of holomorphic functions and their properties.
  • Familiarity with the concept of convergence in series.
  • Knowledge of the unit disc in complex analysis.
  • Basic principles of complex function theory.
NEXT STEPS
  • Study the properties of holomorphic functions in the unit disc.
  • Explore the implications of the Weierstrass factorization theorem.
  • Learn about the distribution of zeros of analytic functions.
  • Investigate the role of boundedness in complex analysis.
USEFUL FOR

Mathematicians, particularly those specializing in complex analysis, students studying holomorphic functions, and researchers interested in the properties of zeros of analytic functions.

Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem.

-----

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
\[\sum_n(1-|z_n|)<\infty.\]

-----

Hint:
Use Jensen's formula.

 
Physics news on Phys.org
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 $\prod_{k=1}^n|z_k|\rightarrow 0$.$\hspace{1in}\blacksquare$
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
3K