Concerning properties of entire functions

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SUMMARY

The discussion revolves around proving that if an entire function \( f: \mathbb{C} \rightarrow \mathbb{C} \) satisfies \( f(z) = z \) for all \( z \) on the circle \( \Gamma \) of radius 1 centered at 0, then \( f(z) = z \) must also hold for all points in the interior of \( \Gamma \). The solution approach involves utilizing the Cauchy Integral Formula, which states that an analytic function is uniquely determined by its boundary values. The participants clarify that the derivative \( f'(z_0) \) cannot be assumed to equal 1 without first establishing that \( f(z) = z \) in the interior.

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  • Understanding of entire functions and their properties
  • Familiarity with the Cauchy Integral Formula
  • Knowledge of complex analysis concepts such as analytic functions
  • Basic proficiency in calculus, specifically differentiation
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  • Study the Cauchy Integral Formula in detail
  • Explore the properties of entire functions and their implications
  • Investigate the concept of analytic continuation
  • Learn about the uniqueness theorem for analytic functions
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Students and professionals in mathematics, particularly those studying complex analysis, as well as educators seeking to clarify the properties of entire functions and their applications in proofs.

Jory
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Homework Statement



let [tex]\Gamma[/tex] [tex]\subset[/tex] C be the circle of radius 1 centred at 0.

Let f: C [tex]\rightarrow[/tex] C be an entire function such that for every z [tex]\in[/tex] [tex]\Gamma[/tex]

f(z) = z

Show that f(z) = z also on Int( [tex]\Gamma[/tex] )

Homework Equations



(f(z) - f(z0) )/(z - z0 ) = f'(z0) perhaps?

The Attempt at a Solution



for z [tex]\in[/tex] Int( [tex]\Gamma[/tex] )

(f(z) - f(z0) )/(z - z0 ) = f'(z0) = 1

=> f(z) = z

This doesn't seem right to me, doesn't really take into account the circular 'boundary' very well.

Any ideas?
 
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No, that doesn't work. To conclude f'(z0)=1 for z0 in the interior using the difference quotient requires you know know f(z)=z in the interior already. And that's what you are trying to prove. Use the Cauchy Integral Formula. An analytic function is determined by its values on the boundary.
 
Of course, thanks
 

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