Why $\frac{p(re^{2\pi is})/p(r)}{|p(re^{2\pi is})/p(r)|}$ defines a loop in the unit circle?

Question

Why can $$\frac{p(re^{2\pi is})/p(r)}{|p(re^{2\pi is})/p(r)|}$$

defines a loop in the unit circle $S^1\subset\mathbb{C}$ based at 1, where $s\in[0,1]$, $p(z)=z^n+a_1z^{n-1}+\ldots+a_n$ is a polynomial for $z\in\mathbb{C}$, $p(z)\neq0$ and $r\in\mathbb{R}$.

I understand the fact that dividing by the magnitude is rescaling it the unit size but what I struggle to grasp is why $p(re^{2\pi is})/p(r)$ is a circle? Is there any way to understand this parametrisation?

I know $re^{2\pi is}$ is parametrisation of a circle with radius $r$, but what does it actually say when we put in the points of the circle into the polynomial $p(z)$?

My other doubt is could $p(re^{2\pi is})/|p(re^{2\pi is})|$ also define a loop in the unit circle? What is the purpose of the $p(r)$ in the original formula?

Thanks for the help.

Answer 1

Let's call your function $f(s)$. For it to be a loop, you just need to have $f(0)=f(1)$. This is true because $e^{2\pi i}=e^{0}=1$, so $f(0)$ and $f(1)$ are both $\frac{p(r)/p(r)}{|p(r)/p(r)|}=1$.