Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear?

This is an exercise in complex analysis:

Let $\Omega\subset{\Bbb C}$ be open and bounded, and $\varphi:\Omega\to\Omega$ a holomorphic function. Prove that if there exists a point $z_0\in\Omega$ such that $$ \varphi(z_0)=z_0\qquad\text{and }\qquad \varphi'(z_0)=1 $$ then $\varphi$ is linear.

I'm trying work out the case $z_0=0$ first, in which $$ \varphi(z)=z+\sum_{n=2}^{\infty}a_nz^2. $$ It suffices to show that $a_n=0$ for all $n\geq 2$. If let $$ \varphi(z)=z+a_2z^2+O(z^3) $$ then $$ \varphi^k(0)=z+ka_2z^2+O(z^3), $$ and $$ \varphi^k(0)=0,\quad (\varphi^k)'(0)=1. $$ If one can show that $\{ka_2\}_{k=1}^{\infty}$ is uniformly bounded, then one at least has $a_2=0$. But I don't know how to go on. Any idea?

Solution 1:

Following your thoughts, when $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n\ge 2$ and $a_n\ne 0$, such that
$$\varphi(z)=z+a_n(z-z_0)^n+O((z-z_0)^{n+1}).$$ As you have noticed, by induction, it follows that for every $k\ge 1$, $$\varphi^k(z)=z+ka_n(z-z_0)^n+O((z-z_0)^{n+1}). \tag{1}$$

Let $r>0$ be such that when $|z-z_0|\le r$, then $z\in\Omega$. Then by $(1)$,

$$ka_n=\frac{1}{2\pi i}\int_{|z-z_0|=r}\frac{\varphi^k(z)}{(z-z_0)^{n+1}}dz.\tag{2}$$ Since $\varphi^k(\Omega)\subset\Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $|\varphi^k|\le M$ on $\Omega$. Then by $(2)$, $$k|a_n|\le Mr^{-n}.$$ Since $k$ is arbitrary, $a_n=0$, a contradiction.