 Solution

The Koebe quarter theorem says that if $F:\mathbb{D}\to \mathbb{C}$ is a univalent function with $F(0)=0$ and $F^{\prime }(0)=1$, thenEquivalently, an arbitrary univalent function $f$ on the unit disc has $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant cannot be increased: the Koebe function $F(w)=w/(1-w{)}^2$ omits the ray $(-\infty ,-1/4]$.