I have the following problem that I am stuck on:
Let $f$ be analytic in $D=\{z\in\mathbb{C}\::\:|z|<1\}$ and suppose that $|f(z)|<M$ for all $z\in D$.
(a) If $f(z_{k})=0$ for $1\leq k\leq n$, show that $$|f(z)|\leq M\prod_{k=1}^{n}\frac{|z-z_{k}|}{|1-\bar{z}_{k}z|}$$ for $|z|<1$.
(b) If $f(z_{k})=0$ for $1\leq k\leq n$, each $z_{k}\neq 0$, and $f(0)=Me^{i\alpha}(z_{1}z_{2}\cdots z_{n})$, find a formula for $f$.
I honestly have no idea of where to begin with this problem. I think that Schwarz's Lemma may be needed somewhere? Thanks in advance for any help!
Function $$g_k(z)=\frac{z-z_k}{1-z\overline{z_k}}$$ has the following properties: a) it has a simple zero at $z_k$, and no other zeros, b) it is analytic in $|z|\leq 1$, and c) $|g_k(z)|=1$ for $|z|=1$. Consider the ratio $$G(z)=\frac{f(z)}{\prod_{k=1}^n g_k(z)} $$ It is analytic (the poles at $z_k$ are removable) and satisfies $|G(z)|\leq M$ for $|z|=1$. By the Maximum Principle $|G(z)|\leq M$ in the unit disk, and this is your inequality.
For b), Maximum Principle says that this inequality is everywhere strict, or everywhere equality. If for example $|G(0)|=M$ then $f=Me^{i\theta}\prod g_k$.