Let $f$ be a holomorphic in $D(0,1)$, with Re$\,f(z) >0$ and $f(0)=1.$ Then $\lvert\, f'(0)\rvert\leq 2$

Question

Let $f:D(0,1) \to \mathbb{C}$ be a holomorphic function, such that

$$ \mathrm{Re} \,f(z) >0\quad \text{and}\quad f(0)=1. $$

How to prove $\lvert\, f'(0)\rvert\leq 2 \ ?$

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This is now a self-answered question.

Answer 1

Hint. Find a Möbius transformation $g$ that takes the right half-plane to the unit disk and $1$ to $0$. Then apply Schwarz's Lemma to $g \circ f$.

Answer 2

Thank you user179549. Let $H$ denote the right plane minus y axis and define $g:H \to D(0,1)$ by $g(z) = \dfrac{z-1}{z+1}.$ It follows that $|g(f(z))| < 1, \forall z \in D(0,1), $ $g \circ f$ is holomorphic in $D(0,1)$ and $g(f(0)) =0.$ By Schwarz lemma, $|g'(f(0))||f'(0)|\leq 1,$ so $|f'(0)| \leq 2.$