Consider a function $f(z)$ defined for all complex $z$ within the open unit circle.
More precisely $$f(z) = a_0 + a_1 z + a_2 z^2 + \cdots$$ where the $a_n$ are complex numbers.
So $f(z)$ is a Taylor series with radius at least $1$ and thus $f(z)$ is analytic within the unit circle.
$f(z)$ is also bounded within the open unit circle.
Now for $|q| = 1$, $f(q)$ is also bounded.
Let $q = \exp(2 \pi t i)$ and thus $f(q) = f(\exp(2 \pi t i))$ for real $t$.
Now we have that $$f(\exp(2 \pi t i)) = \cos(g(t)) + i \sin(g(t)) $$ for a 2pi periodic real $g(t)$.
So it follows that $|f(q)| = 1$
Also we require $g(t)$ to be $C^{\infty}$ over the reals $t$. (This implies that $f(q)$ and $g(t)$ are both fourier series)
Further $f(z) \neq c * id(z)^ k$ for some integer $k$ and some complex $|c| = 1$ or $c=0$.
So far the definition and requirements for $f(z)$.
Now the questions:
What are examples of $f(z)$?
Can the radius of $f(z)$ be larger than $1$?
If the radius is $1$, is it still possible to have analytic continuation? (so the unit circle is not a natural boundary)