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For a conformal bijection $f:D\to \tilde{D}$ between proper planar domains, put $d=\mathrm{dist}(z,\partial D)$ and $\tilde{d}=\mathrm{dist}(f(z),\partial \tilde{D})$. Apply the Koebe quarter theorem to $(f(z+dw)-f(z))/(d{f}^{\prime }(z))$ on the unit disc to obtain $d|f^{\prime }(z)|/4\le \tilde{d}$. Apply it again to $f^{-1}$ to get the opposite bound. Thus $\boxed{\tilde{d}/(4d)\le |f^{\prime }(z)|\le 4\tilde{d}/d}$. The domains need not be simply connected domains. For the whole plane, the boundary distances are infinite and these ratios are undefined.

The map $C(z)={\tan }^2(\pi \sqrt{z}/4)$ is a conformal map from $\{x+iy:y^2<4(1-x)\}$ onto the unit disc. A principal square root first maps the slit parabolic region to a vertical strip, and a scaled tangent followed by squaring maps the strip to a slit disc. Evenness in the square-root variable gives a holomorphic extension across the removed axis, filling the target slit.

The Koebe function $k(z)=z/(1-z{)}^2$ is a univalent function from the unit disc onto $\mathbb{C}\setminus (-\infty ,-1/4]$. To see the image, $(1+z)/(1-z)$ maps the disc onto the right half-plane and $k(z)=(((1+z)/(1-z){)}^2-1)/4$. Its normalization $k(0)=0$, $k^{\prime }(0)=1$ proves sharpness of the Koebe quarter theorem.

If $f$ is a univalent function on the unit disc, then $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant $1/4$ is sharp, as shown by the Koebe function. This gives useful comparisons between the derivative of a conformal map, its conformal radius, and its distance to the domain boundary.

A planar Brownian motion beginning at $a\in (0,1)$ in the unit disc has circular exit law obtained by pulling back uniform angular measure through ${\psi }_a(w)=(w-a)/(1-aw)$. This Möbius transformation sends the starting point to zero. The right semicircle maps to an arc with endpoints at angles $\pm (\pi /2+2\arctan a)$, giving positive-half-plane exit probability $1/2+(2/\pi )\arctan a$.