Proof that an automorphic injective hol. mapping on a bounded domain given one point being the identity implies the whole mapping is too Let $\Omega$ be a bounded domain and let $\phi$ be an injective holomorphic mapping of $\Omega$ to itself. 
The proposition I would like to prove is: 

If $ \exists a \in \Omega$ s.t. $\phi(a) = a$ and $\phi'(a) = 1$,
  $\phi$ must be the identity.

A given hint was: Write the power series for $\phi$ centred at $a$ as:
$$\phi(z)=a+(z-a)+ \text{ higher order terms}$$
and consider $\phi, \phi \circ \phi, \phi \circ \phi \circ \phi, \dots$. Estimate the coefficient of the first nonzero term in the power series after the linear term, assuming that it exists, and show that it must in fact be zero. Unfortunately however, I'm not quite certain how to proceed with this hint.
 A: A translation of to domain $\Omega$ by $a$ and changing $\phi$ to $\phi-a$ clearly doesn't affect this computation at all, so we assume that $a = 0$. This is just a convenience; the computation just looks cleaner this way, and you can probably work it out just fine in the generality you posed by mimicking what occurs below.
$\phi$ has a power series expansion in some ball $B(0,R) \subset \Omega$ whose closure is contained in $\Omega$ as well. It looks like
$$\phi(z) = z + a_nz^n + \mbox{higher order terms} = z +a_nz^n +O(z^{n+1}).$$
Assume that the terms $a_2,\dots, a_{n-1}$ of the power series are all zero. We'll show $a_n$ must be zero as well. Also, I'm using big-O notation for convenience as well; if you don't like it you can just write 'higher order terms' instead and it won't change the reasoning.
To illustrate what is going on we compute the following easy case: 
$$
\begin{align*}
\phi(\phi(z)) &= \phi(z) + a_n(\phi(z))^n + O(z^{n+1}) \\
   &= z + a_n z^n + a_n\left( z + a_nz^n + O(z^{n+1})\right)^n + O(z^{n+1}) \\
   &= z + 2 a_n z^n + O(z^{n+1})
\end{align*}
$$
In general (perhaps proven carefully using induction and the above reasoning) we have 
$$\phi^{(m)}(z) = z + m a_n z^n + O(z^{n+1}).$$
Here, $\phi^{(m)}$ represents the $m$-fold composition of $\phi$.
The Cauchy estimates say that if $f$ is a holomorphic function in a neighborhood of some closed ball $B$ of radius $R$ and any $z$ in that ball we have
$$\left|\frac{d^n}{dz^n}f(z) \right|\leq \frac{n!\sup_{B}|f|}{R^n}.$$
Lets apply the Cauchy estimates to $\phi^{(m)}$ on $B(0,R)$. We can read off the $n$th derivative of this composition from the power series above to see that
$$\frac{d^n}{dz^n} \phi^{(m)}(z) = n!\cdot m \cdot a_n $$
Therefore the Cauchy estimates say that
$$|a_n| \leq\frac{ \sup_{B(0,R)}|\phi^{(m)}(z)|}{m\cdot R^n}$$ 
Since $n$ is fixed, $a_n$ and $R^n$ remain fixed quantities. Since $\phi: \Omega \rightarrow \Omega$, and $\Omega$ is a bounded domain, the supremum  of $|\phi|$, and therefore of $|\phi^{(m)}|$ is uniformily (and most importantly, independently of $m$) bounded by some constant $C$, which corresponds to the norm of the point farthest from the origin in the domain. Rearranging everything, we have 
$$|a_n| \leq \frac{K}{m}$$
where $K$ is some fixed real number independent of $m$. Since $m$ can be chosen arbitrarily large, we have $a_n = 0$. 
We have shown there can be no higher order terms in the power series for $\phi$; $\phi$ must therefore be the identity.
