Koebe's growth theorem

Write $z=r{e}^{i\theta }$. Since $f(0)=0$, $f(z)={\int }_0^r{e}^{i\theta }{f}^{\mathrm{\prime }}(t{e}^{i\theta })dt\mathrm{.}$ Therefore $\mathrm{\mid }f(z)\mathrm{\mid }\le {\int }_0^r\mathrm{\mid }{f}^{\mathrm{\prime }}(t{e}^{i\theta })\mathrm{\mid }dt\mathrm{.}$

For the lower bound, choose $z_0$ on $\mathrm{\mid }z\mathrm{\mid }=r$ for which $\mathrm{\mid }f(z_0)\mathrm{\mid }$ is minimal. The segment from $0$ to $f(z_0)$ lies in $f(\overline{D(0,r)})$: otherwise its first exit point from $f(D(0,r))$ would be an image of the circle $\mathrm{\mid }z\mathrm{\mid }=r$ having modulus strictly smaller than $\mathrm{\mid }f(z_0)\mathrm{\mid }$. Since $f$ is univalent, this segment has a lift $\gamma$ from $0$ to $z_0$.

Applying [L1] inside the integral gives $\mathrm{\mid }f(z)\mathrm{\mid }\le {\int }_0^r\frac{1+t}{(1-t{)}^3}dt=\frac{r}{(1-r{)}^2}\mathrm{.}$

The image of $\gamma$ is a straight segment, so [L1] gives $\mathrm{\mid }f(z_0)\mathrm{\mid }={\int }_{\gamma }\mathrm{\mid }{f}^{\mathrm{\prime }}(\zeta )\mathrm{\mid }\mathrm{\mid }d\zeta \mathrm{\mid }\ge {\int }_{\gamma }\frac{1-\mathrm{\mid }\zeta \mathrm{\mid }}{(1+\mathrm{\mid }\zeta \mathrm{\mid }{)}^3}\mathrm{\mid }d\zeta \mathrm{\mid }\ge {\int }_0^r\frac{1-t}{(1+t{)}^3}dt=\frac{r}{(1+r{)}^2}\mathrm{.}$ The penultimate inequality follows because $\gamma$ joins radius $0$ to radius $r$, while the integrand is positive and depends only on the radius.

Minimality of $z_0$ now gives $\mathrm{\mid }f(z)\mathrm{\mid }\ge \mathrm{\mid }f(z_0)\mathrm{\mid }\ge r\mathrm{/}(1+r{)}^2$ for every $\mathrm{\mid }z\mathrm{\mid }=r$. Together with step 2.1 this proves both bounds.