Concerning properties of entire functions

[Jory]

Homework Statement

let \Gamma \subset C be the circle of radius 1 centred at 0.

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

f(z) = z

Show that f(z) = z also on Int( \Gamma )

Homework Equations

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

The Attempt at a Solution

for z \in Int( \Gamma )

(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?

 

[Dick]

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.

 

[Jory]

Of course, thanks