Riemann mapping theorem

Every nonempty simply connected proper open subset of the complex plane is conformally equivalent to the unit disc.

For a bounded simply connected domain $D$ and a conformal bijection $f:\mathbb{D}\to D$, $f$ extends continuously to the closed disc if and only if $\partial D$ is a locally connected space. A Jordan boundary gives a homeomorphism of closed discs, but local connectedness alone need not give injectivity on the boundary: a slit has two boundary approaches. The same statement applies on the sphere to unbounded domains after a suitable change of coordinates. This boundary theorem is different from the Caratheodory extension theorem for measures.

The Koebe distortion theorem gives universal upper and lower bounds on a univalent function and its derivative in terms of the distance to the boundary. In particular, derivatives at two points whose hyperbolic distance is bounded differ by at most a universal multiplicative factor.

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.

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.