Let $f$ be an analytic function on the closed unit disk $\overline{\mathbb{D}}$. On its boundary $\partial \mathbb{D}$ it holds that $\lvert\,f(z) -z\rvert < \lvert z\rvert$.
I now have to show that $$ \left\lvert\,\, f'\left(\frac{1}{2}\right)\right\rvert \leq 8. $$
I already figured out, that there cannot be a $\,z_0 \in \partial \mathbb{D}$, such that $\,f(z_0) =0,\,$ since that would mean that $\lvert 0-z_0\rvert < \lvert z_0\rvert\,$ which produces a contradiction, since the inequality is strict.
I also know, that $f$ takes its maximum on $\partial \mathbb{D}$ according to the maximum modulus principle.
My assumption is that i should get $\,\lvert\,f'(z)\rvert < \lvert z^{-3}\rvert\,$ by looking at the numbers, which seem a bit random to me.
But now I'm stuck. Any help would be greatly appreciated!