This question was already asked here, I wouldn’t ask it again however I do not think that the question gets the point across that it’s trying to get across, and I can’t find anything else related to this question online.
The question is found in Donald E. Marshall’s Complex Analysis
3.8 (a) Prove that $\phi$ is a one-to-one analytic map of $\mathbb{D}$ onto $\mathbb{D}$ if and only if $$ \phi(z) = c \left( \frac{z - a}{1 - \overline{a}z} \right), $$ for some constants $c$ and $a$, with $|c| = 1$ and $|a| < 1$. What is the inverse map?
(b) Let $f$ be analytic in $\mathbb{D}$ and satisfy $|f(z)| \to 1$ as $|z| \to 1$. Prove that $f$ is rational.
and the point that I think the question is trying to get across is, can one show (b) given that (a) is proven? Otherwise why would Donald bunch these two questions together?
I was able to prove (a) using the Schwarz lemma and automorphisms of the disk and I’m thinking it’s related to (b) since $|\phi(z)|\to 1$ as $|z|\to 1$ since $|c|=1$, however any further and I’m stuck.
I’m looking for a solution that builds on (a) if possible.
Any help is appreciated!
I guess this is technically an answer (?).
Conrad's answer here is probably what you were looking for; you simply needed to know what the word "Blaschke product" means. Given a finite collection of complex numbers $\{a_1, a_2, ..., a_n\}$, a Blaschke product is a rational function of the form $$B(z) = K \prod _{j = 1}^n \dfrac{z - a_j}{1 - \overline{a_j}z}, \quad |K| = 1.$$ If you like you can even match the notation of (a) and write $$B(z) = \phi_1(z) \phi_2 (z) \cdots \phi_n(z), \quad \phi_j(z) = c_j\left(\dfrac{z - a_j}{1 - \overline{a_j}z}\right), \quad K = c_1c_2 \cdots c_n$$ That is, $B(z)$ is a product of functions of the form in part (a). As Conrad suggested in their answer, having proven (a) you know you can form a Blaschke product whose zeros exactly agree with the finitely many zeros of $f(z)$.
The solution doesn't require you to know anything about Blaschke products or what they are called.