Baby Rudin 7.21: Explanation of proof I cannot understand the hint of the problem below...
Could someone possibly explain it to me??

Let $K$ be the unit circle in the complex plane (i.e., the set of all z with $\mid z \mid=1$), and let $\mathscr{A}$ be the algebra of all functions of the form
$f(e^{i\theta})=\sum_{n=0}^N c_n e^{in\theta}, \ (\theta \ \ \ real)$.
Then $\mathscr{A}$ separates points on $K$ and $\mathscr{A}$ vanishes at no point of $K$, but nevertheless there are continuous functions on $K$, which are not in the uniform closure of $\mathscr{A}$.
Hint: For every $f\in\mathscr{A}$
$ \int_0^{2\pi} f(e^{i\theta})e^{i\theta}d\theta=0$
and this is also true for every $f$ in the closure of $\mathscr{A}$.

Also, in the solution manual of this problem, it is stated that $\mathscr{A}$ separating points and not vanishing on $\mathscr{A}$ is obvious.
Could you please explaing why this is the case as well?
I am a beginner in functional analysis (especially in cases involving complex domains), so any comments would be greatly appreciated.
Thank you.
https://minds.wisconsin.edu/bitstream/handle/1793/67009/rudin%20ch%207.pdf?sequence=5&isAllowed=y
 A: $f(z)=z$ is a function on $\mathscr A$ that satisfies
$1).\ $ Whenever  $z_1\neq z_2\in S^1,\ f(z_1)\neq f(z_2)$ (this is trivial).
$2).\ $ For all $z\in S^1,\ f(z)=z\neq 0$ (trivial also because $0\notin S^1$.
This answers the first part of the question.
Now, if $f\in \overline{\mathscr A}$ then there is a sequence of functions $(f_n)\subseteq \mathscr A$ such that $f_n\to f\ \textit{uniformly}$ and this implies that $\ f_n(\theta)e^{i\theta}\to f(\theta)e^{i\theta}$ uniformly as well. Since the convergence is uniform, it follows that $\int^{2\pi}_0\ f_n(\theta)e^{i\theta}d\theta\to \int^{2\pi}_0f(\theta)e^{i\theta}d\theta.$ Now, the left hand side of this is, with $f_n(\theta)=\sum^N_{k=0}c_{nk}e^{ik\theta},\ \sum^N_{k=0}c_{nk}\int^{2\pi}_0e^{i(k+1)\theta}d\theta=0.$ Therefore, $\int^{2\pi}_0f(\theta)e^{i\theta}d\theta=0.$
So, to finish, we need a continuous function on the circle for which $\int^{2\pi}_0f(\theta)e^{i\theta}d\theta\neq 0.$ If you think about it, the obvious choice is $f(\theta)=e^{-i\theta}$ because $f$ is continuous on the circle, and since $\mathscr A$ consists of all linear combinations of $\{e^{in\theta}\}$ and since $e^{-i\theta}$ is not such a combination (because $\{e^{in\theta},e^{-i\theta}\}$ is a linearly idependent set), so our $f$ is not a linear combination of elements of $\mathscr A$ and so we suspect that it can't be of the form $\sum^{\infty}_{n=1}c_ne^{in\theta}$ either. To prove it, just check that $\int^{2\pi}_0e^{-i\theta}e^{i\theta}d\theta=2\pi\neq 0,$ so $f\neq \overline{\mathscr A}$.
