Koebe's distortion theorem

By rotating the source and target, it is enough to treat the case $z=r\in [0,1)$. Define ${\psi }_r(\zeta ):=\frac{\zeta +r}{1+r\zeta },g(\zeta ):=\frac{f({\psi }_r(\zeta ))-f(r)}{(1-r^2)f^{\mathrm{\prime }}(r)}\mathrm{.}$ Fact [L1] makes ${\psi }_r$ an automorphism of $\mathbb{D}$, so $g\in \mathcal{S}$.

Differentiate twice at $0$. Since ${\psi }_r^{\mathrm{\prime }}(0)=1-r^2$ and ${\psi }_r^{\mathrm{\prime }\mathrm{\prime }}(0)=-2r(1-r^2)$, one gets $g^{\mathrm{\prime }\mathrm{\prime }}(0)=(1-r^2)\frac{f^{\mathrm{\prime }\mathrm{\prime }}(r)}{f^{\mathrm{\prime }}(r)}-2r\mathrm{.}$ Because $g\in \mathcal{S}$, [L2] gives $\mathrm{\mid }{g}^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\mid }\le 4$. Therefore $\mid \frac{f^{\mathrm{\prime }\mathrm{\prime }}(r)}{f^{\mathrm{\prime }}(r)}-\frac{2r}{1-r^2}\mid \le \frac{4}{1-r^2}\mathrm{.}$

Put $H(r):=\log \text{\hspace{0.17em}⁣}((1-r^2)f^{\mathrm{\prime }}(r)),$ choosing a continuous branch along $[0,r]$ since $f^{\mathrm{\prime }}$ never vanishes on $\mathbb{D}$ for univalent $f$. Then step 2.1 yields $\mathrm{\mid }{H}^{\mathrm{\prime }}(t)\mathrm{\mid }=\mid \frac{f^{\mathrm{\prime }\mathrm{\prime }}(t)}{f^{\mathrm{\prime }}(t)}-\frac{2t}{1-t^2}\mid \le \frac{4}{1-t^2}(0\le t<1)\mathrm{.}$

Integrating step 3.1 from $0$ to $r$ and using $f^{\mathrm{\prime }}(0)=1$ gives $\mathrm{\mid }H(r)\mathrm{\mid }\le {\int }_0^r\frac{4}{1-t^2}dt=2\log \frac{1+r}{1-r}\mathrm{.}$ Exponentiating and dividing by $1-r^2$ yields $\frac{1-r}{(1+r{)}^3}\le \mathrm{\mid }{f}^{\mathrm{\prime }}(r)\mathrm{\mid }\le \frac{1+r}{(1-r{)}^3}\mathrm{.}$

The argument of steps 1.1 through 4.1 applies after rotation to every point of modulus $r$, so the same bounds hold for the original $z$.