Schwarz lemma type bound for holomorphic function on unit disk

Question

I encountered the following old qualifier problem:

Suppose $f: \mathbb{D} \to \mathbb{D}$ is holomorphic, $f(0)=0$, and $f'(0)=1$. Show that $$|f(z)-z| \leq 2|z|^2 \, \forall z \in \mathbb{D}. \, \, \, \, \, (1)$$

I believe I can solve this by applying the Schwarz lemma to the function $$g(z)=\frac{f(z)}{2z}-\frac{1}{2}.$$ What seems a little strange is that from the beginning, the Schwarz lemma tells us exactly what $f(z)$ looks like: namely, it is a rotation. Is it possible to prove (1) straight from the fact that $f(z)=az$ for some $|a|=1$?

Answer 1

In fact, it is trivial, since the assumption $f'(0)=1$ implies $a=1$, so $f(z) \equiv z$.