Solving a Complex Problem: Proving a Function Reduces to a Polynomial

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SUMMARY

The discussion focuses on proving that a function \( f(z) \), which is analytic across the entire complex plane and satisfies the inequality \( |f(z)| < |z|^n \) for some integer \( n \) and sufficiently large \( |z| \), must reduce to a polynomial. The key approach involves expressing \( f(z) \) as a power series \( f(z) = \sum_{m=0}^{\infty} a_m z^{m} \) and demonstrating that the condition leads to \( a_m = 0 \) for all \( m \geq N \) for some natural number \( N \). The application of the Cauchy Integral Formula is crucial in establishing that the derivatives \( f^{(n+m)}(a) = 0 \) for any complex number \( a \) and \( m \geq 1 \).

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  • Familiarity with power series representation of functions.
  • Knowledge of Cauchy Integral Formula and its implications.
  • Basic grasp of Liouville's theorem in complex analysis.
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  • Study the implications of Liouville's theorem in complex analysis.
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  • Explore the concept of entire functions and their growth rates.
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AlbertEinstein
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Hi all.

The problem is "Prove that a function which is analytic in the whole plane and satisfies an inequality |f(z)| < |z|^n for some n and sufficiently large |z| reduces to a polynomial." I do not understand what I need to show that the function reduces to a polynomial.

Any help will be appreciated.
Thanks.
 
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A more general theorem holds,that no entire function dominates another.. Check out :
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis )
 
Last edited by a moderator:
AlbertEinstein said:
Hi all.

The problem is "Prove that a function which is analytic in the whole plane and satisfies an inequality |f(z)| < |z|^n for some n and sufficiently large |z| reduces to a polynomial." I do not understand what I need to show that the function reduces to a polynomial.

Any help will be appreciated.
Thanks.

Write f(z) = \sum_{m=0}^{\infty} a_m\ z^{m}

(which you can do since it's entire).

Show that |f(z)| < |z|^n implies for some natural number N,

a_m = 0

for any m >= N.
 
Last edited:
You can apply Cauchy Integral Formula to show that f^(n+m)(a) = 0 for any complex number a in C and m>=1.
 

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