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Let $\phi_1, \phi_2 : \mathbb{D} \to \mathbb{C}$ be two injective holomorphic mappings such that $\phi_1(0) = \phi_2(0) =0$ and $\phi_1'(0) = \phi_2'(0)$. Is it true that $\phi_1 \equiv \phi_2$? More generally, is $\phi_1 \equiv \phi_2$ if the codomain is $\mathbb{C}^n$?

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Let $\phi_1(z)=z$, and let $\phi_2=\phi_1+f$, where $f$ is any function with $f(0)=f'(0)=0$ and $|f'(z)|<1/2$ for all $z$.

It's clear that $\phi_2$ is injective: $$|\phi_2(z)-\phi_2(w)|\ge|z-w|-|f(z)-f(w)|\ge\frac12|z-w|.$$

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Let $\phi_1$ be any non-linear such mapping (e.g., $\phi_1(z)=(1-z)^2$) and $\phi_2(z)=\phi_1(0)+\phi_1'(0)z$.

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    $\begingroup$ $\phi_1$ and $\phi_2$ need to be injective. $\endgroup$ – Jaikrishnan Jan 3 '18 at 11:35
  • $\begingroup$ @Jaikrishnan My example $\phi_1$ is injective on $\Bbb D$ (and that's why I demnaded it to be "such" a mapping), my $\phi_2$ is injective unless $\phi_1'(0)=0$ (which is not the case for the specific $\phi_1$ I named) $\endgroup$ – Hagen von Eitzen Jan 4 '18 at 18:43

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