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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?

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    $\begingroup$ If can you reduce to the case of $f(0)=0$, for instance as in Akhil's answer, then Schwarz lemma is useful. $\endgroup$ – user1119 Dec 8 '10 at 20:08
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One can reduce it to the case of $f(0)=0$ (i.e. $0$ is the fixed point) by making a suitable linear fractional transformation of the disk. Namely, if $z^*$ is the fixed point, apply the above argument to $L \circ f \circ L^{-1}$ where $L$ sends $z^* \to 0$ and is a LFT.

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  • $\begingroup$ Is there a simpler way to reduce to the case of $f(0)=0$? Just judging from the part of the textbook this is taken from, it is supposed to be solved without the knowledge of Mobius transformations. $\endgroup$ – jake Dec 8 '10 at 21:45
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    $\begingroup$ @Jake: Not that I see offhand. I would say that Mobius transformations, however, are simpler than --- say --- Rouche's theorem or Schwarz's lemma. $\endgroup$ – Akhil Mathew Dec 8 '10 at 22:03

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