For any complex polynomial $p(z)$ of order $m$, we showed earlier that on a circle $S$ of sufficiently large radius $r$ in the plane, $$\frac{p(z)}{|p(z)|}\quad \text{and}\quad \frac{z^m}{|z^m|}=\left(\frac{z}{r}\right)^m$$ are homotopic maps of $S \to S^1$. Thus $p/|p|$ must have the same degree as $(z/r)^m$ - namely, $m$. When $m>0$ we conclude $p/|p|$ will not extend to the whole disk of radius $r$, implying that $p$ must have a zero inside the disk.
So my questions are:
What is $S$?
Why not extending to the whole disk implies have a zero?
As written, $S$ is "a circle of sufficiently large radius $r$".
If $p$ had no zero, then $z\mapsto \frac{p(z)}{|p(z)|}$ would define a map from the disk bounded by $S$ to the circle $S^1$