Let $f$ be analytic on $mathbb{D}$, $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$. Prove $|f'(0)| leq 4$.
1
$begingroup$
I am trying to solve the following problem: Let $f$ be analytic on $mathbb{D} = {z in mathbb{C}: |z|<1}$ , $f(0)=-1$ and $|1+f(z)| < 1+|f(z)|$ for $|z|<1$ . Prove $|f'(0)| leq 4$ . I have an idea of how to do it, but I am stuck at one of the steps. I'd like to use Schwarz' lemma. So I wish to find a function $h = g circ f, $ such that $h(0) = 0$ , and where $g: f(mathbb{D}) rightarrow mathbb{D}$ . Then $h$ would be a function from $mathbb{D}$ to $mathbb{D}$ , and Schwarz' lemma would apply, so that $|h'(0)| = |g'(f(0))f'(0)| leq 1.$ If I have then managed to find a $g$ such that $g'(-1) = 1/4$ , I believe I would be done with the proof. Right now, however, I am stuck on what $f(mathbb{D})$ is. I suppose I should use the assumption that $|1+f(z