A graduate level question in complex analysis

Click For Summary
SUMMARY

The discussion centers on proving that if f and g are entire functions satisfying the condition |f(z)| ≤ |g(z)| for all z in the complex plane, then f must equal cg for some complex constant c. The user attempts to apply Liouville's theorem to the function f/g, noting that f/g is bounded and analytic except at the zeros of g. The key challenge is to demonstrate that f/g remains analytic at the zeros of g, which would allow the application of Liouville's theorem to conclude that f/g is constant.

PREREQUISITES
  • Understanding of entire functions in complex analysis
  • Familiarity with Liouville's theorem
  • Knowledge of analytic functions and their properties
  • Concept of zeros of functions and their multiplicities
NEXT STEPS
  • Study the implications of Liouville's theorem in complex analysis
  • Research the properties of entire functions and their zeros
  • Learn about the concept of analytic continuation
  • Explore the relationship between bounded functions and their limits in complex analysis
USEFUL FOR

Graduate students in mathematics, particularly those specializing in complex analysis, as well as educators and researchers seeking to deepen their understanding of entire functions and Liouville's theorem.

vijigeeths
Messages
4
Reaction score
0
If f and g are two entire functions such that mod(f(z)) <= mod(g(z)) for all z in C, prove that f=cg for some complex constant c.
 
Physics news on Phys.org
I tried to prove this by applying Liouville's theorem to f/g.
it is clear that f/g is bounded.
and f/g is analytic except at the zeros of g.
if z1 is a zero of g its a zero of f also.
and if multiplicity of z1 as a zero of g will be less than or equal to multiplicity of z1 as a zero of f.
i got stuck here.
pls help me to prove f/g is analytic at zeros of g also.
so that by Liouville's theorem f/g will be a constant.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K