The identity theroem complex analysis

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SUMMARY

The discussion centers on proving the non-existence of a holomorphic function \( f \) in the open unit disk such that \( f(1/n) = \frac{(-1)^n}{n^2} \) for \( n = 2, 3, 4, \ldots \). The identity theorem is referenced, stating that if two holomorphic functions agree on a set with a limit point, they must be identical. Previous examples established that no holomorphic function can satisfy similar conditions, specifically \( f(1/n) = 2^{-n} \). The proposed function \( g(z) = (-1)^{\frac{1}{z}} z^2 \) is suggested as a potential tool for analysis.

PREREQUISITES
  • Understanding of holomorphic functions
  • Familiarity with the identity theorem in complex analysis
  • Knowledge of limit points in topology
  • Basic concepts of sequences and convergence in complex analysis
NEXT STEPS
  • Study the implications of the identity theorem in complex analysis
  • Explore examples of holomorphic functions and their properties
  • Investigate the behavior of sequences in the context of complex functions
  • Examine the function \( g(z) = (-1)^{\frac{1}{z}} z^2 \) and its holomorphic characteristics
USEFUL FOR

Students and researchers in complex analysis, particularly those studying holomorphic functions and the identity theorem, will benefit from this discussion.

Scousergirl
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Homework Statement


Prove that there is no holomorphic function f in the open unit disk such that f(1/n)=((-1)^n)/(n^2) for n=2,3,4...


Homework Equations


The identity theorem: Let f and g be holomorphic functions in the connected open subset of C, G. If f(z)=g(z) for all z in a subset of G that has a limit point in G, then f=g.



The Attempt at a Solution



We proved in an earlier example that there is not holomorphic function f in the open unit disk such that f(1/n)=2^(-n) n=2,3,4... Is the above function identical to this? I thought this was the route to go but i am not sure anymore. I know we have to somehow use the identity theorem.
 
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Maybe it helps to consider the function [itex]g(z)=(-1)^{\frac{1}{z}}z^2[/itex]
 

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