Koebe function ($k(z)=z/(1-z{)}^2$)

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.