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This is exercise 9 of chapter III.4 in Freitag and Busam.

$\mathbb{E}:=\{z\in\mathbb{C} | |z|<1\}$. Let $f,g:\mathbb{E}\rightarrow \mathbb{E}$ be bijective analytic functions, which satisfy $f(0)=g(0)$ and $f'(0)=g'(0)$. Moreover, assume that $f'$ and $g'$ have no common zero. Show that $f(z)=g(z)$ for all $z\in\mathbb{E}$.

The indication was to use Lemma III.3.8 for the function $f\circ g^{-1}$. The lemma states: Let $\varphi:\mathbb{E}\rightarrow \mathbb{E}$ be a bijective map with zero as a fixed point, such that both $\varphi$ and $\varphi^{-1}$ are analytic. Then there exists a complex number $\zeta$ with modulus $1$ such that $$\varphi(z)=\zeta z.$$

My question is, how do we know if $f\circ g^{-1}$ is analytic and has an analytic inverse? Is $f^{-1}$ analytic? Why do we need to assume that $f'$ and $g'$ have no common zero?

Jiu
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  • While reading through this question: If $f(z)=g(z)$ and both are analytic, then $f'(z)=g'(z)$ and if one of the two functions has a zero, the other one has to have one as well. What do you mean by "common" zero? Do you mean a simple zero (a zero of degree 1)? – F. Conrad Feb 06 '18 at 14:23
  • @F.Conrad We need to show that $f=g$, it's not in the hypothesis. – Jiu Feb 06 '18 at 14:27
  • I know, but assuming it is true, then $f'$ and $g'$ have indeed common zeroes. It is a contradiciton. – F. Conrad Feb 06 '18 at 14:37
  • @F.Conrad Ok I see what you mean. It would simply imply that $f'$ does not have any zero. – Jiu Feb 06 '18 at 14:40
  • For the inverse being analytic: https://math.stackexchange.com/questions/1048204/the-inverse-of-a-bijective-holomorphic-function-is-also-holomorphic – Arnaud Mortier Feb 06 '18 at 14:43

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Use this answer to see why $f^{-1}$ should be analytic (and in the way, why $f'$ and $g'$ never vanish anyway).

Then, after applying the lemma to $f^{-1}\circ g$, you have $g'(0)=\zeta f'(0)$, this is where you need $f'(0)$ and $g'(0)$ to not be zero, so that $\zeta =1$.

Arnaud Mortier
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