Bloch's Theorem. Let $f$ be an analytic function on a region containing $\overline{\mathbb{D}}$ and satisfying $f(0)=0$, $f'(0)=1$. Then there is a disk $S\subset \mathbb{D}$ on wich $f$ is one-one and such that $f(S)$ contains a disk of radius $\frac{1}{72}$.
Suppose that in the statement of Bloch's Theorem it is only assumed that $f$ is analytic on $\mathbb{D}$. What conclusion can be drawn ?
(Hint: Consider the functions $f_{s}(z)=s^{-1}f(sz)$, $0<s<1$.)
For every $s$, $f_s$ is defined and analytic on a disk $\mathbb{D}_s$ with radius $1/s$ (bigger than the unit disk), and it satisfies $f_s(0) = 0$ (obvious) and $f_s'(0)=1$ (apply chain rule). Then, every one of the $f_s$ verifies the hypothesis of Bloch's theorem, so they are injective on (maximal) disks $S_s$, with $f_s(S_s)$ containing a disk of radius $1/72$.
Since $f_s$ is injective for $z\in S_s$, then $f$ must be injective whenever $z$ is in $sS_s$, which is a smaller disk. You can see that the maximal disk $D$ of injectivity of $f$ determines all $S_s = \frac{1}{s}D$.
We know that $f_s(S_s)$ contains a disk of radius $1/72$, but $f_s(S_s) = f_s(\frac{1}{s}D) =\frac{1}{s}f(D)$. Since $\frac{1}{s}f(D)$ contains a disk of radius $1/72$ for all $s\in (0,1)$, we deduce that $f(D)$ contains also a disk of radius $1/72$.
We deduce that $f$ is one-to-one on a disk $D$ and that $f(D)$ contains a disk of radius $1/72$. That is, Bloch's theorem is still true.