Suppose $f: \mathbb{D} \to \mathbb{D}$ is a holomorphic function and $f(0)=0$. The function $f$ has a total of five zeros (counting multiplicities) in the closed half-disc $\overline{\frac{1}{2}\mathbb{D}} = \{z \in \mathbb{C}: |z| \leq \frac{1}{2}\}$. How large can $|f'(0)|$ be?
Schwarz Lemma $f: \mathbb{D} \to \mathbb{C}$ is analytic with $f(0) = 0$ and $|f(z)| \leq 1$ on $\mathbb{D}$, then $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$ and $|f'(0)| \leq 1$
It would be almost indisputable that Schwarz lemma will be of use here. Now, given that $f$ has five zeros in the domain $\overline{\frac{1}{2}\mathbb{D}}$, it's no question that Rouche's theorem will likely also be of use here.
Since $f$ maps $\mathbb{D}$ to itself, then implicitly we also have $|f(z)| \leq 1$ for $z$ on $\mathbb{D}$. Then $f$ satisfies Schwarz's lemma, and so an upper bound for $|f'(0)|$ is $1$. Perhaps this bound can be improved.
But I have not used the fact that $f$ has five zeros in $\overline{\frac{1}{2}\mathbb{D}}$. Again, I know this must be Rouche's theorem, but I'm not sure how to apply it.
Best Answer
Let $a_1, \ldots, a_4$ denote the four zeros of $f$ with $0 < |a_k| \le 1/2$. Then $$ f(z) = h(z) \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z} $$ with $h: \Bbb D \to \Bbb D$ and $h(0) = 0$. It follows that $$ f'(0) = h'(0) \prod_{k=1}^n (-a_k) \, . $$ We have $|h'(0)| \le 1$ by the Schwarz lemma, so that $$ |f'(0)| \le \prod_{k=1}^n |a_k| \le \frac{1}{2^4} = \frac{1}{16} \, . $$ The bound is sharp, equality holds if and only if $$ f(z) = c z \prod_{k=1}^n \frac{z-a_k}{1-\overline{a_k} z} $$ with complex numbers $c, a_1, \ldots, a_4$ satisfying $|c|=1$ and $|a_1|=|a_2|=|a_3|=|a_4| = 1/2$.