Koebe quarter theorem

Remarks

After replacing $f$ by $(f - f(0))/f'(0)$, one may assume $f(z) = z + a_2 z^2 + \cdots$ is in the class $S$. If $w \notin f(\mathbb{D})$, then $g = wf/(w - f)$ is again in $S$, and its second coefficient is $a_2 + 1/w$. Bieberbach's inequality $\abs{a_2} \le 2$ for functions in $S$, which follows from the Grönwall area theorem, gives $\abs{a_2 + 1/w} \le 2$ and $\abs{a_2} \le 2$, hence $\abs{1/w} \le 4$, that is, $\abs{w} \ge 1/4$. The constant $1/4$ is sharp: the Koebe function $k(z) = z/(1-z)^2$ is in $S$ and omits the value $-1/4$. Injectivity is necessary: $f(z) = (e^{Nz} - 1)/N$ has $f(0) = 0$ and $f'(0) = 1$, but omits the value $-1/N$, which is close to $0$ for large $N$.

Going deeper

For a student

The theorem is the basic covering theorem of univalent function theory. It gives the distance estimate $\frac{1}{4}\abs{f'(z)}(1 - \abs{z}^2) \le \operatorname{dist}(f(z), \partial f(\mathbb{D})) \le \abs{f'(z)}(1 - \abs{z}^2)$ for univalent $f$, where the upper bound comes from the Schwarz lemma. This compares the hyperbolic metric of a simply connected domain with the reciprocal of the distance to its boundary, up to a factor $4$. Together with the Koebe distortion theorem, it gives the compactness of the class $S$ in the topology of locally uniform convergence. For maps that need not be injective, Bloch's theorem gives a much weaker covering result.