= Solution
Assume the <domains> are proper, so $d$ and $\widetilde d$ are finite and positive, and interpret the <conformal transformation> as a bijection. Apply the <Koebe quarter theorem> to
$$
F(w)=\frac{f(z+dw)-f(z)}{d f'(z)},\qquad |w|<1.
$$
The <open ball> $B(z,d)$ lies in $D$, so $F$ is a normalized <univalent function>. Its image inclusion shows that $B(\widetilde z,d|f'(z)|/4)\subseteq\widetilde D$. Therefore $d|f'(z)|/4\leq\widetilde d$, or $|f'(z)|\leq4\widetilde d/d$.
Apply the same argument to the inverse <conformal map>, whose <derivative> at $\widetilde z$ is $1/f'(z)$. It gives $\widetilde d/(4|f'(z)|)\leq d$. Thus the <boundary-distance derivative bound for a conformal bijection> is
$$
\boxed{\frac{\widetilde d}{4d}\leq|f'(z)|\leq\frac{4\widetilde d}{d}.}
$$
The 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, $D=\widetilde D=\mathbb C$ 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 $d=\widetilde d=1$ at zero but $|f'(0)|=1/8<1/4$. Thus an interpretation allowing a map into, rather than onto, the second <domain> makes the lower bound false.
Back to article page
