Holomorphic function on a connected bounded open subset of the complex plane
Problem. Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_0 \in \Omega$ such that $$ \varphi\left(z_0\right)=z_0 \quad \text { and } \quad \varphi^{\prime}\left(z_0\right)=1 $$ then $\varphi$ is linear.
This exercise (Exercise 9 in Chapter 2) from Complex Analysis by SteinShakarchi is intended to be solved using Cauchy's Theorem or its corollaries presented in the chapter.
However, the authors omitted the assumption that $\Omega$ is connected. This assumption is necessary, as a counterexample can easily be constructed by defining $\varphi(z) = z^2$ on the connected components that do not contain $z_0$.
The textbook provides a critical hint for solving the problem:
Hint: Why can one assume that $z_0=0$ ? Write $\varphi(z)=z+a_n z^n+O\left(z^{n+1}\right)$ near 0 , and prove that if $\varphi_k=\varphi \circ \cdots \circ \varphi$ (where $\varphi$ appears $k$ times), then $\varphi_k(z)=$ $z+k a_n z^n+O\left(z^{n+1}\right)$. Apply the Cauchy inequalities and let $k \rightarrow \infty$ to conclude the proof. Here we use the standard $O$ notation, where $f(z)=O(g(z))$ as $z \rightarrow 0$ means that $f(z) \leq Cg(z)$ for some constant $C$ as $z \rightarrow 0$.
Comments on the hint:

The reduction to assuming $z_0 = 0$ is a fairly standard step for experienced readers but may appear mysterious to beginners. Understanding the details of "WLOG" (Without Loss of Generality) part in various arguments is rewarding.

The hint mentions the standard big O notation in its asymptotic form. In number theory, a nonasymptotic form, where the inequality $f(z) \le Cg(z)$ holds for all $z$ in the domain, is frequently used.
I will write my solution to the problem below. Alternative approaches or broader insights beyond the original question are also welcome.
1 answer
First, as noted in the problem statement, we should explicitly assume that $\Omega$ is connected.
Now, let’s proceed with the reduction suggested in the hint, specifically assuming $z_0 = 0$.
If $z_0\ne 0$, we can consider the holomorphic function $\psi: \tilde{\Omega} \to \tilde{\Omega}$ where $$\psi(z) = \varphi(z + z_0)  z_0,$$ with $\tilde{\Omega} = \Omega  z_0 := \{z  z_0 : z \in \Omega\}$, which shifts the original domain. This transformation ensures that $\psi(0) = 0$ and $\psi'(0) = 1$. If we can show that $\psi$ is linear, i.e., $\psi(z) = z$, given the conditions at $0$, it follows that $\varphi(z) = \psi(z  z_0) + z_0 = z$, which is the desired result.
For beginners, it's important to avoid sloppy reasoning that does not properly extend the conclusion of the general case from the special case.
Another useful reduction is through analytic continuation, which allows us to restrict our proof to the neighborhood of $0$. If we can show that $\varphi(z) = z$ near $0$, the result holds for the entire domain.
Given the conditions, we can express $\varphi(z)$ near $0$ as: $$ \varphi(z) = z + a_n z^n + O(z^{n+1}), $$ where $a_n$ is the coefficient of the first nonzero higherorder term. We aim to show that $a_n = 0$, implying there can be no higherorder terms in the power series.
By induction, as suggested in the hint, we find: $$ \varphi_k(z) = z + k a_n z^n + O(z^{n+1}). $$
Assume this expansion holds in $\overline{D(0,R)}\subset \Omega$ where $D(0,R)$ is the open disc centred at $0$ with radius $R$.
Using the relation between the coefficients of a power series and its derivatives, along with Cauchy's inequalities, we can estimate $k a_n$ in terms of the norm of $\varphi_k$:
$$ k a_n = \left\frac{\varphi_k^{(n)}(0)}{n!}\right \le \frac{\\varphi_k\_{\partial D(0, R)}}{R^n} \le \sup_{z \in \Omega} \varphi(z) / R^n\ . $$Since this holds uniformly in $k$, we conclude that $a_n = 0$, thereby completing the proof.
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