A condition implying that a holomorphic function is the identity map Let $\Omega$ be an open, connected and bounded subset of $\mathbb{C}$, and $\varphi : \Omega \rightarrow \Omega$ a holomorphic function.
If there exists a $z_{0} \in \Omega$, such that
$$
\varphi(z_0) = z_{0}\quad\text{and}\quad\varphi'(z_0) = 1,
$$ 
then prove that $\varphi$ is the identity function, 
i.e. $\varphi(z)=z$, for all $z\in\Omega$.
Any hints?
 A: As $\Omega$ is bounded, then so is $\varphi$, i.e., $|\varphi(z)|\le M$, for some $M>0$.
Without loss of generality assume that $z_0=0$. If $\varphi$ is not the identity map, then near $z=0$ 
$$
\varphi(z)=z+a_kz^k+{\mathcal O} (z^{k+1}),
$$
where $a_k$ is the first nonzero coefficient of its expansion. Clearly, if $\varphi^{n\circ}=\underbrace{\varphi\circ\cdots\circ\varphi}_{\text{$n$ times}}$, then $\big|\varphi^{n\circ}(z)\big|\le M$, and
$$
\varphi^{n\circ}(z)=z+na_kz^k+{\mathcal O} (z^{k+1}).
$$
This means that 
$$
\frac{d^k}{dz^k}\big(\varphi^{n\circ}(z)\big)_{z=0}=n k!a_k
$$
As $0\in\Omega$, there is an $r>0$, such that $\bar D_r\subset\Omega$, where $\bar D_r$ is the closed disc of radius $r$ and centered at the origin. According to Cauchy integral formula
$$
\frac{d^k}{dz^k}\big(\varphi^{n\circ}(z)\big)_{z=0}=\frac{k!}{2\pi i}\int_{|z|=r}\frac{\varphi^{n\circ}(z)\,dz}{z^{k+1}},
$$
and hence
$$
n\,|a_k| \le \frac{M}{r^k},
$$
for every $n\in\mathbb N$, which is a contradiction, as the left hand side tends to infinity, when $n\to\infty$ and the right hand side is constant.
