I am having difficulties with the following problem:
$\bf Given$: $f $ is analytic and maps from unit disk to itself. $\bf Prove:$ $|f'(0)|\leq1- |f(0)|^2 $.
For some reason (unclear to me) it is hinted to consider: $h(z) = \frac{f(z)-\alpha}{\alpha^* f(z)-1}$ and to let $\alpha =f(0)$.
(note: $\alpha^*$=complex conjugate of $\alpha$.)
I have absolutely no clue how to do this. Can someone please help?
We know that for any $w$ with $\lvert w\rvert < 1$, the map
$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}\cdot z}$$
is an automorphism of the unit disk $\mathbb{D}$.
So if $f \colon \mathbb{D} \to \mathbb{D}$ is a holomorphic mapping, then $h = T_{f(0)} \circ f$ also is a holomorphic map from the unit disk to itself. And we have $h(0) = 0$ by construction. So $h$ is a map satisfying the preises of Schwarz' lemma, whence we have
$$\lvert h(z)\rvert \leqslant \lvert z\rvert\tag{1}$$
for all $z\in\mathbb{D}$. Now we can look at the form of $h$, as far as we know it. Expanding $T_{f(0)}$, the inequality $(1)$ becomes
$$\left\lvert \frac{f(z)-f(0)}{1 - \overline{f(0)}\cdot f(z)}\right\rvert \leqslant \lvert z\rvert \tag{2}$$
for all $z\in\mathbb{D}$. Rearranging $(2)$ yields
$$\left\lvert \frac{f(z)-f(0)}{z}\right\rvert \leqslant \lvert 1 - \overline{f(0)}\cdot f(z)\rvert.$$
The last step should not be difficult to find.