A question on the maximizing the value of $|f(0)|$ given that $f(z)$ is analytic for $|z|<3$

Question

Let $D=\{ z \in \mathbb{C}: |z|<3\}.$ Suppose that $f:D\rightarrow\mathbb{C}$ is an analytic function such that $|f(z)|<1$ for all $z \in D.$ Additionally, $f(\pm1)=f(\pm i)=0.$ If this is the case, then what is the maximmum value of $|f(0)|?$ For which functions is the maximum value attained ?

$\textbf{My attempt}:$ Since it is given that $f(\pm1)=f(\pm i)=0,$ we need $x^4-1$ as a factor of $f(z).$ This is the observation I am able to make. I am not able to proceed further. Any ideas would be highly helpful.

Answer 1

In general, the main idea involved are disks centered at the origin, so either Schwarz's lemma or Blaschke products are usually expected. Here are some heuristics:

1. By the hypothesis in the question we have $f \colon D_3(0) \to D_1(0)$ so that $f(0) \in D_1(0)$, then the map $F \colon D_1(0) \to D_1(0)$ defined by $F(z):=f(3z), \forall z\in D_1(0)$ satisfies $F(\frac{1}{3}) = f(1) = 0.$
2. Define $B \colon D_1(0) \to \mathbb{C}$ by $$B(z) = \frac{1/3 - z}{1 - (1/3)z}\cdot \frac{-1/3 - z}{1 - (-1/3)z}\cdot \frac{i - z}{1 + i z} \cdot \frac{-i - z}{1 - iz}$$ it can be easily checked that $B$ is analytic in $B_1(0)$ and for all $|z| = 1$ we have $|B(z)| = 1$ and by maximum modulus principle $B \colon D_1(0) \to D_1(0)$.
3. Consider the composite function $G \colon D_1(0) \to D_1(0)$ by $G(z) = \frac{F(z)}{B(z)}$ then $G$ is clearly analytic in $D_1(0)$ and for any $0 < r < 1$, applying maximum modulus principle to $G$ on $B_r(0)$ so that $$\max_{z\in \overline{D_r(0)}}|g(z)| = \max_{|z| = r}|g(z)| = \max_{|z| = r} \left| \frac{F(z)}{B(z)} \right| = \max_{\theta \in [0, 2\pi]} \left| \frac{F(re^{i \theta})}{B(re^{i \theta})}\right| \to 1$$ as $r \to1^-$ so we have the following bound $$|F(z)| = |f(3z)| \leq \left|\frac{1/3 - z}{1 - (1/3)z}\cdot \frac{-1/3 - z}{1 - (-1/3)z}\cdot \frac{i - z}{1 + i z} \cdot \frac{-i - z}{1 - iz} \right| = |B(z)|$$ for all $z \in D_1(0)$. To attain the bound, set $z =0$ and we get $$|f(0)| \leq \left| \frac{1}{3} \cdot \frac{-1}{3} \cdot \frac{i}{3} \cdot \frac{-i}{3} \right| = \frac{1}{81}.$$ To attain equality, that is to attain the maximum modulus at some point interior to $D_1(0)$ so the function $G$ is a constant function that takes the value $|G(z)| = 1$ for all $z \in D_1(0)$. Hence, there exists $\theta \in \mathbb{R}$ so that $f(3z) = B(z) e^{i\theta}$, and so $$f(z) = e^{i\theta}\cdot \frac{1/3 - (z/3)}{1 - (1/3)(z/3)}\cdot \frac{-1/3 - (z/3)}{1 - (-1/3)(z/3)}\cdot \frac{i - (z/3)}{1 + i (z/3)} \cdot \frac{-i - (z/3)}{1 - i(z/3)}$$ for all $z \in D_3(0)$.