Let $Ω$ be a bounded domain in $\mathbb{C}$ and φ be a conformal mapping of $Ω$ to itself. Let $P ∈ Ω $ and suppose both that $φ(P ) = P$ and $φ′(P ) = 1.$
Prove that φ must be the identity.
Here's what I have: In my textbook the hint was to write the power series $φ(z) = P + (z − P) + · · ·$ and consider the sequence $φ, φ ◦ φ, φ ◦ φ ◦ φ, . . . . $ Apply Cauchy ’ s estimates to the first nonzero coefficient of the power series for $φ$ after the term $(z −P)$. At first, my thought was to consider a family of functions of the form $\{ φ^{k}(z) \}_{k = 1}^{\infty}$ and I think this sequence has a subsequence that converges uniformly to a holomorphic function, $f$ on $Ω$ by Montel's Theorem and then I can show that $|f'(z)| \geq |φ^{k}(z)|$ for all z in $Ω$.
The other way to approach this is by using Schwartz Lemma but I am not sure if that is the right way though. Can you guys only give me some hints on how to solve this problem? Try not to work it out completely please. Thank you very much!!!
Using normal families you can argue as follows: If $\varphi$ is not the identity then $$ \varphi(z) = z + (z-P) + a_n (z-P)^n + O((z-P)^{n+1}) $$ for $z \to P$ and some $n\ge 2$, $a_n \ne 0$. Use induction to show that the $k^\text{th}$ iterate satisfies $$ \varphi^{k}(z) = z + (z-P) + k a_n (z-P)^n + O((z-P)^{n+1}) $$
As you correctly said, the sequence $\{ \varphi^{k} \}_{k = 1}^{\infty}$ is bounded, and has a subsequence $\{ \varphi^{k_l} \}_{l = 1}^{\infty}$ which converges locally uniformly to holomorphic function $f$ in $D$.
Now use that locally uniform convergence is inherited by derivatives to derive a contradiction.