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 $\left| z \right|<R$, then $\left| a_{ n } \right| \le r^{-n}\underset { \left| z \right| =R }{ \sup } \left| f(x) \right| $.

  • 1
    $\begingroup$ 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 $$ |a_n|\leq r^{-n} \sup_{|z|=r}|f(z)| $$ $\endgroup$
    – sharding4
    Jun 30, 2017 at 16:38
  • $\begingroup$ You're right. I'll add that to the question. Thanks! $\endgroup$
    – Richard
    Jun 30, 2017 at 16:47

1 Answer 1


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.


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