Schwarz lemma in Stein book Show that if $f\colon D(0,R)→\mathbb{C}$  is holomorphic with $|f(z)|\leq M$ for some $M>0$ then 
$$\bigg|
\frac{f(z)-f(0)}{M^2-\bar{f(0)}f(z)}\bigg|\leq \frac{|z|}{MR}
$$
This is an exercise of Stein's book Schwarz lemma section. But I do not have any idea how can I solve it, first at all, the Schwarz lemma needs that the origin is fixed by the function, but I do not have any value, only that it is bounded. So have I to do a conformal map first?
I would be grateful if you give any idea.
 A: Ok so note that since $|f(z)|<M$ you can actually re-write: $f:D_0(R)\to \overline{D_0(M)}$. However by the maximum modulus principle we know that f doesn't achieve its supremum on the open set $D_0(R)$ and thus we actually have: 
$$f:D_0(R)\to D_0(M)$$
Now by the Riemann Mapping Theorem (although honestly it's not really necessary to see what i'm about to do) we know that every simply connected open subset of $\mathbb{C}$ can be mapped conformally to the unit disc $\mathbb{D}$. That gives us the existence of the following two maps:
$$\varphi : D_0(R)\to\mathbb{D} \quad \text{and} \quad \pi : D_0(M)\to\mathbb{D}$$
Both of which are conformal maps defined by $z\mapsto z/R$ and $z\mapsto z/M$ repectively. 
Now we define the function:
$$ F =  \pi\circ f\circ \varphi^{-1} : \mathbb{D}\to\mathbb{D}$$
Note that by the chain rule we have that $F$ is holomorphic and so we have the following inequality (following from the Schwarz-Pick theorem letting $z_1 = z$ and $z_2 = 0$) [Shoutout to my boi @Dominic Petti for pointing this out]:
$$\left|\frac{F(z) - F(0)}{1 -\overline{F(0)}F(z)}\right|\leq |z| \implies 
\left|\frac{\pi\circ f(zR) - \pi\circ f(0)}
{1 -\overline{\pi\circ f(0)}\pi\circ f(zR)}\right|\leq |z|$$
Since we define $\pi(z) = z/M$,
$$ \implies \left|\frac{\frac{1}{M}(f(zR) - f(0))}
{1 -\frac{1}{M^2}\overline{f(0)}f(zR)}\right|\leq |z|
\implies M \left|\frac{f(zR) - f(0)}{M^2 -\overline{f(0)}f(zR)}\right|\leq |z|$$
Now letting $\zeta = zR$ and thus $|z| = |\zeta|/R$ we obtain the desired inequality: 
$$ \left|\frac{f(\zeta) - f(0)}{M^2 -\overline{f(0)}f(\zeta)}\right|\leq \frac{|\zeta|}{MR}$$
A crispy result.
