Application of Liouville's Theorem (Complex Analysis)

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SUMMARY

The discussion centers on the application of Liouville's Theorem to prove that a meromorphic function \( f:\mathbb{C}\rightarrow\mathbb{C} \) satisfying the condition \( |f(z)|^5 \leq |z|^6 \) for all \( z \in \mathbb{C} \) must be identically zero. The attempt to apply Liouville's Theorem to the quotient \( \frac{f(z)^5}{z^6} \) reveals that while this function is bounded, it must also be entire. The conclusion drawn is that \( f(z) \) must have a zero at the origin, leading to the assertion that \( f(z) \) is either identically zero or takes the form \( f(z) = z^n g(z) \) with \( n > 1 \).

PREREQUISITES
  • Understanding of Liouville's Theorem in complex analysis
  • Familiarity with meromorphic functions
  • Knowledge of entire functions and their properties
  • Basic skills in manipulating complex functions and limits
NEXT STEPS
  • Study the implications of Liouville's Theorem on bounded entire functions
  • Research the properties of meromorphic functions in complex analysis
  • Explore examples of entire functions and their zeros
  • Learn about the relationship between zeros of functions and their growth rates
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Students and professionals in mathematics, particularly those focusing on complex analysis, as well as educators looking for examples of Liouville's Theorem applications.

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Homework Statement


Prove that if [tex]f[/tex] is a meromorphic function [tex]f:\mathbb{C}\rightarrow\mathbb{C}[/tex] with

[tex]|f(z)|^5\leq |z|^6\quad\textrm{for all}\quad z\in\mathbb{C}[/tex]

Then [tex]f(z)=0[/tex] for all [tex]z\in\mathbb{C}[/tex]

Homework Equations


Liouville's Theorem

A bounded entire function is constant.

The Attempt at a Solution


I tried applying Liouville's theorem to the quotient [tex]f(z)^5/z^6[/tex] which is bounded by 1 but was unsuccessful in proving that f is constant.
 
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Note that the hypothesis implies that [tex]f(z)^5/z^6[/tex] is a bounded function but because of the [tex]z^6[/tex] in the denominator, you must prove that it's also entire.

To prove this, note (again, by the hypothesis) that [tex]f(z)[/tex] must have a zero at the origin, so either [tex]f(z)[/tex] is identically zero, or [tex]f(z) = z^ng(z)[/tex]; but then

[tex]f(z)^5/z^6=z^{5n}g(z)^5/z^6[/tex]

And this implies that [tex]n > 1[/tex]. After simplifying, what can you say about [tex]g(z)[/tex]?
 

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