Suppose that $f(z)$ is complex analytic on $|z| \leq 1$ and satisfies $|f(z)| < 1$ for $|z|=1$.
(a) Prove that the equation $f(z)=z$ has exactly one root (counting multiplicities) in $|z|<1$.
(b) Prove that if $|z_0| \leq 1$, then the sequence $z_n$ defined recursively by $z_n= f(z_{n-1}) , n=1,2,...$, converges to the fixed point of $f$.
I was able to prove (a) using Rouche's theorem, but (b) stumps me. I know that (b) is true for analytic fuctions such that $f(0)=0$ or $|f'(z)|<1$ on the disc, neither of which are necessarily true in general. The farthest I was able to get was $|f(z)-z^*|<\frac{1}{1-|z*|}|z-z^*|$, where $z^*$ is the fixed point of $f$, but $\frac{1}{1-|z^*|}>1$, so I don't think this helps me. Can someone please point me in the right direction?