Consequences of a fixed point of a bounded, complex analytic function 
Let $G$ be an open and bounded region of $\mathbb{C}$, and let $f$ be analytic on $G$ so that $f(G) \subset G$ and, for some $a\in G$, $f(a) = a$ and $|f'(a)| = r$.
(a) If $r < 1$, then $f^n \to a$, ($f^n$ converges to the constant function $a$ in the space of analytic functions) where $f^n := f\circ f\circ \cdots f$ ($n$-times)
(b) If $f'(a) = 1$, then $f(z) = z$ for all $z \in G$.

For part (a), I have done the following: the family $\{f^n\}$ is normal by Montel's theorem, so for any subsequence $\{f^{n_k}\}$ there is a subsequence $\{f^{n_{k_s}}\}$ which converges to some analytic function $g$.
Further, we know that $(f^n)'(a) = (f')^n(a) = r^n \to 0$ as $n \to \infty$.  Beyond that, I'm stuck.
For part (b), I'm thinking that it is sufficient to prove the statement for when $a = 0 \in G$, and to show that the Taylor series for $f(z) - z$ must be identically 0.  But I'm also stuck here.
 A: We must assume that $G$ is connected, otherwise the statement is wrong since  $f$ can be defined independently in each connected component of $G$.
Without loss of generality we can assume that $a=0$ to simplify the notation. 

Part (a)  We have $f(0) = 0$ and $|f'(0)| = r < 1$. It follows that there are $c \in [0, 1)$ and $\delta > 0$ such that $B(0, \delta) \subset G$ and
$$
  |f(z)| \le c |z| \quad \text{for } z \in B(0, \delta) \, ,
$$
i.e. $f$ is a contraction on a sufficiently small disk. Then
$$
   |f^n(z)| \le c^n |z| \quad \text{for } z \in B(0, \delta)
$$
holds for all iterates $f^n$, so that $f^n \to 0$ uniformly in $B(0, \delta)$.
As you already noticed, $\{ f^n \}$ is a normal family in $G$, so that every subsequence $ f^{n_k} $ has a locally uniformly convergent subsequence $ f^{n_{k_j}} \to g$. As shown above, $g(z)=0$ for $z \in B(0, \delta)$, so that $g=0$ in $G$ because of the identity theorem.
So $g=0$ is the only possible limit function of any convergent subsequence of $f^n$, which implies that $f^n \to 0$ locally uniformly in $G$.

Part (b) Here we have $f(0) = 0$ and $f'(0) = 1$. Let us assume that $f$ is not the identity function and derive a contradiction. We have
$$
  f(z) = z + cz^{m} + O(z^{m+1}) \quad \text{for } z \to 0
$$
for some $m \ge 2$ and some $c \ne 0$, and consequently
$$
  f^n(z) = z + ncz^{m} + O(z^{m+1}) \quad \text{for } z \to 0
$$
for each iterate $ f^n$. In particular, the $m$-th derivative of $f^n$ at $z=0$ is equal to $nc$, which shows that no subsequence of $f^n$ can converge in a neighborhood of $z=0$. This is a contradiction since $\{f^n\}$ is a normal family in $G$.
(The proof of the second part is essentially taken from “§7 Parabolic Fixed Points” in Milnor.)
