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If I have two analytic complex functions $f_1$ and $f_2$ which each conformally map the unit disc to regions $\Omega_1$ and $\Omega_2$ respectively, where $\Omega_1\subset\Omega_2$, such that $f_1(0)=f_2(0)$, what can I say about the relationship between the moduli of the derivatives at $0$? It seems like it'd be necessary for $|f_2'(0)|$ to be larger to compensate for the difference in range size. Is it possible for that relationship to switch somewhere else in the disc (i.e. $|f_2'(z_0)|\le|f_1'(z_0)|$ for some $z_0\ne 0$ in the unit disc) while still preserving the containment configuration of the ranges? I suspect that conformality prevents that from being possible, and that it may be the mechanism which allows the initial growth rate of the functions to dictate their future growth.

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  1. Yes, $|f_1'(0)|\le |f_2'(0)|$ because $f_2^{-1}\circ f_1$ satisfies the assumptions of the Schwarz lemma.
  2. Yes, the inequality may be reversed elsewhere in the disk. For example, $f_1(z)=z$ and $f_2(z)=(1+z)^2-1$ satisfy the above assumptions but $|f_2'(z)|=2|1+z|$ is less than $|f_1'(z)|=1$ when $z=-2/3$.
  3. Despite 2, the inequality $|f_1'(z_0)|\le |f_2'(z_0)|$ still holds at those points $z_0\ne 0$ where $f_1(z_0)=f_2(z_0)$. The proof is again an application of the Schwarz Lemma, but in a more general (Schwarz-Pick) form.
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