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Let $\Delta$ denote the open unit disc.

Let $G$ be a simply connected region and $G\neq\mathbb{C}$. Suppose $f:G\rightarrow\Delta$ is a one-to-one holomorphic map with $f(a)=0$ and $f'(a)>0$ for some $a$ in $G$. Let $g$ be any other holomorphic, one-to-one map of $G$ onto $\Delta$. Express $g$ in terms of $f$.

Attempt:

Set $\alpha=g(a)$ and $$\phi_\alpha=\frac{z-\alpha}{1-\overline{\alpha}z}\in\operatorname{Aut}\Delta.$$ Then $f\circ g^{-1}\circ \phi_\alpha^{-1}$ is an automorphism of the unit disc which fixes 0. Hence $f\circ g^{-1}\circ \phi_\alpha^{-1}(z)=e^{i\theta}z$ for some $\theta\in[0,2\pi)$. Therefore $g(z)=\phi^{-1}(e^{-i\theta}f(z))$.

Question:

Since I haven't used the fact that $f'(a)>0$ (and I know that this makes $f$ unique), is there a way to be more specific about the form of $g$ using this information?

Thanks in advance.

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This looks good to me. –  froggie May 11 '12 at 19:00

1 Answer 1

up vote 1 down vote accepted

Your solution works for any fixed $f$ whether or not $f'(a)>0$. The condition $f'(a)>0$ is just a convenient way to fix the phase of the derivative; this affects the exact value of $\theta$ in your argument, but not its existence or uniqueness.

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I see. Thanks!! –  John Adamski May 12 '12 at 3:13

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