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Assume the domains are proper, so $d$ and $\tilde{d}$ are finite and positive, and interpret the conformal transformation as a bijection. Apply the Koebe quarter theorem toThe open ball $B(z,d)$ lies in $D$, so $F$ is a normalized univalent function. Its image inclusion shows that $B(\tilde{z},d|f^{\prime }(z)|/4)\subseteq \tilde{D}$. Therefore $d|f^{\prime }(z)|/4\le \tilde{d}$, or $|f^{\prime }(z)|\le 4\tilde{d}/d$.

Apply the same argument to the inverse conformal map, whose derivative at $\tilde{z}$ is $1/f^{\prime }(z)$. It gives $\tilde{d}/(4|f^{\prime }(z)|)\le d$. Thus the boundary-distance derivative bound for a conformal bijection isThe full domains need not be simply connected domains: only the two interior discs are used. If the whole plane is allowed as a domain, a conformal bijection onto another plane domain is affine and both domains are the whole plane. For example, $D=\tilde{D}=\mathbb{C}$ and $f(z)=z$ make both boundary distances infinite and the printed ratios $\infty /\infty$ undefined. This is a necessary proper-domain convention. Bijectivity also matters: the injective conformal map $f(w)=w/8$ from the unit disc into itself has $d=\tilde{d}=1$ at zero but $|f^{\prime }(0)|=1/8<1/4$. Thus an interpretation allowing a map into, rather than onto, the second domain makes the lower bound false.