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I have an analytic function $f$ mapping the (unit) disc to the (unit) disc, with $f(0)=b$ in $D$, and $f(b)=0$.

Part (a) was to show that $|f'(b)f'(0)|\leq 1$, which I have already done by applying the Schwarz lemma to $f(f(z))$.

Next, part (b) states that

Suppose $f'(b)f'(0)=1$. Find all possible functions $f$.

I am stuck.

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up vote 3 down vote accepted

The version of the Schwarz Lemma I know says that for holomorphic $g:D\rightarrow D$ , if in addition to $g(0)=0$ you have $|g(c)| = |c| $ for at least one $c\neq 0$ or $|g'(0)|=1$, then $g$ is a rotation.

You may apply the latter again to $g(z)=f\circ f(z)$, i.e. $g'(0)=f'(f(0))f'(0)= f'(b)f'(0)=1$. So g(x)=ax for some complex number $a$ s.t. $|a|=1$, in this case clearly $a=1$. (Your heading seems to indicate you figured that out already, but I'm not sure)

Consequently, $f$ is onto and one-one in $D$ (why?), hence $f(z)$ is a biholomorphic map $D\rightarrow D$. These are well known, and I assume you know them, too, since this is what you usually prove using the Schwarz lemma and the reason to introduce the lemma. They are of the form $$f(z)= \frac{az+b}{\bar{a}z + \bar{b}} \,\,, a,b \in \mathbb{C}, |a|^2 -|b|^2 =1$$ If you know that the rest is just calculation.

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Why exactly is f 1-1 and onto? – Frank White Jul 26 '12 at 20:01
@AllenCox because $f\circ f$ is. $f(z)=f(w)\Rightarrow f(f(z))=f(f(w))\Rightarrow z=w$ so $f$ is 1-1, and if $w=f\circ f(z)$ then $w=f(x)$ for $x=f(z)\in D$, so $f$ is onto. – user20266 Jul 27 '12 at 5:47

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