# Stuck on Exercise in Stein and Sharkarchi Complex Analysis

I'm currently self-studying Stein and Sharkarchi's complex analysis book, and I'm stuck on the following exercise from chapter 2:

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

$${\varphi}(z_0)=z_0$$ and $${\varphi}'(z_0)=1$$

then $$\varphi$$ is linear.

The following hint is provided:

Why can one assume that $$z_0 = 0$$? Write $${\varphi}(z) = z + a_nz^n + O(z^{n+1})$$ near $$0$$, and prove that if $${\varphi}_k = \varphi \circ\cdots\circ \varphi$$ (where $$\varphi$$ appears $$k$$ times), then $${\varphi}_k(z) = z + ka_nz^n + O(z^{n+1})$$. Apply the Cauchy inequalities and let $$k \to\infty$$ to conclude the proof.

I've verified that one can indeed assume that $$z_0=0$$ and that $${\varphi}_k(z) = z + ka_nz^n + O(z^{n+1})$$, but I'm totally at a loss as to how the Cauchy inequalities are relevant here. Any help would be greatly appreciated!

EDIT: The Cauchy inequality referenced here is that if $$f(z)=\sum _{ n=0 }^{ \infty }{ a_nz^n }$$ is holomorphic in $$\left| z \right|, then $$\left| a_{ n } \right| \le r^{-n}\underset { \left| z \right| =R }{ \sup } \left| f(x) \right|$$.

• The Cauchy inequality referred to is probably this one According to Wikipedia: if $f(z)=\sum a_nz^n$ is holomorphic in $|z|<R$ and $0<r<R$ then the coefficients satisfy Cauchy's inequality $$|a_n|\leq r^{-n} \sup_{|z|=r}|f(z)|$$ Jun 30, 2017 at 16:38
• You're right. I'll add that to the question. Thanks! Jun 30, 2017 at 16:47

Don't forget that $\Omega$ is bounded. Let $K\in(0,+\infty)$ be such that $(\forall z\in\Omega):|z|\leqslant K$. Then, since $\varphi$ is a function from $\Omega$ into itself, then $(\forall z\in\Omega):\bigl|\varphi(z)\bigr|\leqslant K$ and, more generally,$$(\forall k\in\mathbb{N})(\forall z\in\Omega):\bigl|\varphi_k(z)\bigr|\leqslant K.$$Now, apply Cauchy's inequality to $\varphi_k$. You will get $k|a_n|\leqslant r^{-n}K$. But, if $a_n\neq0$, then $\lim_{k\in\mathbb N}k|a_n|=+\infty$, whereas $r^{-n}K$ is a fixed number.