Extending function analytically

Question

I'm trying to understand why the following is true: Suppose you have a complex holomorphic function such that $f(0)=0$, $0<|f'(0)|=\lambda<1$. Then there exists $r>0$ such that the sequence $\phi_n(z)=\frac{f^n(z)}{\lambda^n}$ ($f^n = f\circ\dots\circ f$) converges uniformly to an analytic function $\phi$ in the disk $\mathbb{D}(0,r)$. The proof of that I understand. My problem is with the following: If $A=\{z|f^n(z)\rightarrow0\}$, then we can extend $\phi$ to $A$ by setting $\phi(z)=\lim_{n\rightarrow \infty}\frac{f^n(z)}{\lambda^n}$. Although I see the functions must converge pointwise in $A$ since for each $z\in A$, $f^n (z)\in \mathbb{D}(0,r)$ for $n$ sufficiently large (thus the sequence is Cauchy), I don't see why $\phi$ would be analytic in $A$, since we can only guarantee uniform convergence in the disk... Is there another way to argument that this extension is analytic?