Bound on the derivative of a holomorphic function at $0$ Suppose $f(z)$ is holomorphic in $|z|< 1$ and $\operatorname{Re}f(z) > 0$ in this region. Furthermore, suppose $f(0) = a > 0$. Then why does $|f'(0)| \leq 2a$?
The hint I have is to consider $g(z) = \frac{f(z) - a}{f(z) + a}$ and then show $|g'(0)| \leq 1$. If I do have $|g'(0)| \leq 1$, then the result follows as $|f'(0)| = 2a|g'(0)| \leq 2a$, but I have no idea on how to show $|g'(0)| \leq 1$.

It follows from the Schwarz Lemma after you observe that
$$T \colon w \mapsto \frac{w-a}{w+a}$$
is a Möbius transformation that maps the right half plane to the unit disk with $T(f(0)) = T(a) = 0$.
The Schwarz Lemma says that for a holomorphic $f \colon\mathbb{D}\to\mathbb{D}$ with $f(0)= 0$, you have $\lvert f'(0)\rvert \leqslant 1$.
You can easily derive that from the Cauchy integral for the derivative,
$$f'(0) = \frac{1}{2\pi i} \int_{\lvert \zeta\rvert = r} \frac{f(\zeta)}{\zeta^2}\,d\zeta,$$
for which the standard estimate yields $\lvert f'(0)\rvert \leqslant \frac{1}{r}$. Letting $r \to 1$ yields $\lvert f'(0)\rvert \leqslant 1$.
