Question

Prove that if f is analytic at z 0 then f in conformal

Answer 1

I am reading over Rudin's discussion of conformal mappings in "Real and Complex Analysis." Rudin states that "no analytic function preserves angles at any point where its derivative is zero. We omit the easy proof of this." So I am trying to fill in the proof.

Suppose Ω⊆C is open and connnected, and that we are given z0∈C and f:Ω→C. Futhermore, suppose that there is some punctured disk D(z0,r)∖{z0}⊆Ω on which f(z)≠f(z0). Then Rudin defines f to be angle preserving at z0 iff

limr→0e−iθf(z0+reiθ)−f(z0)|f(z0+reiθ)−f(z0)|

exists and is independent of θ.

So, with the additional assumption that f′(z0)=0, I want to suppose that the limit above exists (and is independent of θ), and then arrive at some contradiction.

Hints or solutions are greatly appreciated.