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.
Derivative of Complex Conformal Maps
2
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complex-analysis
1 Answers
1
- Yes, $|f_1'(0)|\le |f_2'(0)|$ because $f_2^{-1}\circ f_1$ satisfies the assumptions of the Schwarz lemma.
- 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 $z=-2/3$.
- 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.