This thought arose after reading J.P May's "A Concise Course in Algebraic Topology" (page 11).
Let $f\colon \mathbb{C} \to \mathbb{C}$ be the map $f(x) = x^n + c_{n-1}x^{n-1} + \dots + c_1x + c_0$. Assume $f(x) \neq 0$ for $|x| \geq 1$. Let $\alpha(x,t) = t^nf(x/t) = x^n + t(c_{n-1}x^{n-1} + \dots + t^{n-1}c_0)$. Then the map $G\colon S^1\times I \to S^1$ defined by $G(x,t) = \alpha(x,t)/|\alpha(x,t)|$ is a homotopy between $f(x)/|f(x)|$ and $x^n$.
On the other hand, define $\beta(x,t) = c_{n-1}x^{n-1} + t(x^n + c_{n-2}x^{n-2} + \dots + 1)$. Define $H\colon S^1\times I \to S^1$ by $H(x,t) = \beta(x,t)/|\beta(x,t)|$. Then $H$ looks like a homotopy between $f(x)/|f(x)|$ and $c_{n-1}x^{n-1}$. But if this were a homotopy, we would have that the degree (in the topological sense) of $f(x)/|f(x)|$ must be both $n$ and $n-1$, which is obviously not the case.
My assumption is that $H$ is not continuous, and that it has something to do with $x^n$ being the dominating term. Is there an easy way to see that this is the case?
You are assuming that $f(x) \neq 0$ for $|x| = 1$. But this should in no way guarantee that either $\alpha(x,t)$ or $\beta(x,t)$ are not zero on the unit circle for all $t$, and so I don't see how the homotopies are necessarily well defined.