Prove f=0: Holomorphic Polynomial

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Discussion Overview

The discussion centers around proving that a holomorphic polynomial \( f \) is identically zero under the condition that the integral of \( f(z) \bar{z}^j \) over the boundary of a unit disk is zero for all non-negative integers \( j \). The scope includes mathematical reasoning and exploration of complex analysis concepts.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant states the integral condition and asks for a proof that \( f \equiv 0 \).
  • Another participant suggests using a polynomial of order greater than \( j \) and explores the implications of the integral being zero, questioning their understanding of the problem.
  • A different participant proposes that if the integral around a boundary is zero, then the function must be analytic, leading to the conclusion that \( f \) must be zero due to the dependence on a non-analytic conjugate function.
  • One participant inquires about the applicability of Green's theorem to the problem.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants present differing approaches and interpretations of the integral condition and its implications for \( f \).

Contextual Notes

Participants express uncertainty about assumptions that can be made regarding the properties of holomorphic functions and the implications of the integral condition.

cheeez
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f is a holomorphic polynomial and if

[tex]$\oint_{\partial D(0,1)}f(z)\bar{z}^{j}dz=$ 0 for j = 0,1,2,3...[/tex]

where [tex]$\partial D(0,1)$[/tex] is the boundary of a disc of radius 1 centered at 0

prove f [tex]$\equiv$[/tex]0
 
Last edited:
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any ideas? if we use any polynomial of order > j like j+1 so f(z) = z^(j+1)

then f(z)(z_bar)^j becomes z(z z_bar)^j = z |z|^2j which is just z since modulus is 1 on boundary of disk of radius 1. so obviously integral of z is 0 since its holomorphic but f(z) is not trivial. am i misunderstanding this.
 
Last edited:
no one?
 
Does it happen that the degree of f is j or greater than j?

What I would do is appeal to the fact that if the integral around a boundary of a simple jordan curve is 0 then the function must be analytic (holomorphic), and then since you have an analytic function times a conjugate function that is still analytic it must be zero

Maybe better phrased as follows:
since the integral around the circle is zero the function being integrated is analytic
so
the analytic function depends on the conjugate function (a non analytic function), so if you take the "d-bar" differential (the dereivative with respec to the congugate of z) you should get zero, which will imply that f must be zero

This might not be as clear as you want, but I'm not sure what you're allowed to assume.
 
have you tried green's theorem?
 

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