Koebe function
= Koebe function
{c}
{title2=$k(z)=z/(1-z)^2$}
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 $k(0)=0$, $k'(0)=1$ proves sharpness of the <Koebe quarter theorem>.